problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
7.185. $6-\left(1+4 \cdot 9^{4-2 \log _{\sqrt{3}} 3}\right) \log _{7} x=\log _{x} 7$. | ## Solution.
Domain of definition: $0<x \neq 1$.
Switch to base 7. We have $6-\left(1+4 \cdot 9^{0}\right) \log _{7} x=\frac{1}{\log _{7} x} \Leftrightarrow$ $\Leftrightarrow 5 \log _{7}^{2} x-6 \log _{7} x+1=0$. Solving this equation as a quadratic in terms of $\log _{7} x$, we get $\left(\log _{7} x\right)_{1}=\fra... | \sqrt[5]{7};7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,360 |
7.186. $\log _{12}\left(4^{3 x}+3 x-9\right)=3 x-x \log _{12} 27$. | ## Solution.
Domain of definition: $4^{3 x}+3 x-9>0$.
Rewrite the equation as
$$
\log _{12}\left(4^{3 x}+3 x-9\right)+\log _{12} 27^{x}=3 x \Rightarrow \log _{12} 27^{x}\left(4^{3 x}+3 x-9\right)=3 x
$$
from which $27^{x}\left(4^{3 x}+3 x-9\right)=12^{3 x} \Leftrightarrow 4^{3 x}+3 x-9=4^{3 x}, 3 x-9=0, x=3$.
Answ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,361 |
7.187. $x^{2} \cdot \log _{x} 27 \cdot \log _{9} x=x+4$. | ## Solution.
Domain of definition: $0<x \neq 1$.
Switch to base 3, then
$$
\frac{3 x^{2}}{\log _{3} x} \cdot \frac{\log _{3} x}{2}=x+4 \Leftrightarrow 3 x^{2}-2 x-8=0
$$
from which $x_{1}=2, x_{2}=-\frac{4}{3} ; x_{2}=-\frac{4}{3}$ does not satisfy the domain of definition.
Answer: 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,362 |
7.188. $\sqrt{\log _{5}^{2} x+\log _{x}^{2} 5+2}=2.5$. | ## Solution.
Domain of definition: $00, \\ \log _{5}^{2} x-2.5 \log _{5} x+1=0\end{array} \Rightarrow\left(\log _{5} x\right)_{3}=\frac{1}{2}>0,\left(\log _{5} x\right)_{4}=2>0\right.$, from which $x_{3}=\sqrt{5}, x_{4}=25$.
Answer: $\frac{1}{25} ; \frac{1}{\sqrt{5}} ; \sqrt{5} ; 25$. | \frac{1}{25};\frac{1}{\sqrt{5}};\sqrt{5};25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,363 |
7.189. $\log _{x} m \cdot \log _{\sqrt{m}} \frac{m}{\sqrt{2 m-x}}=1$. | Solution.
Domain of definition: $\left\{\begin{array}{l}0<m \neq 1, \\ 0<x \neq 1, \\ x<2 m .\end{array}\right.$
Switch to base $m$, then
$\frac{1}{\log _{m} x} \cdot \frac{\log _{m} \frac{m}{\sqrt{2 m-x}}}{\log _{m} \sqrt{m}}=1 \Leftrightarrow \log _{m} x+\log _{m}(2 m-x)=2 \Rightarrow$
$\Rightarrow \log _{m} x(2 ... | m | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,364 |
7.190. $\log _{2} 3+2 \log _{4} x=x^{\frac{\log _{9} 16}{\log _{3} x}}$. | ## Solution.
Domain of definition: $0<x \neq 1$.
From the condition, we have $\log _{2} 3+\log _{2} x=x^{\frac{\log _{3} 4}{\log _{3} x}} \Leftrightarrow \log _{2} 3+\log _{2} x=x^{\log _{x} 4} \Rightarrow$ $\Rightarrow \log _{2} 3+\log _{2} x=4, \log _{2} 3 x=4$, from which $3 x=16, x=\frac{16}{3}$.
Answer: $\frac{... | \frac{16}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,365 |
7.191. $\log _{10} x+\log _{\sqrt{10}} x+\log _{\sqrt[3]{10}} x+\ldots+\log _{\sqrt[1]{10}} x=5.5$. | Solution.
Domain of definition: $x>0$.
Let's switch to base 10. We have
$$
\lg x+2 \lg x+3 \lg x+\ldots+10 \lg x=5.5, (1+2+3+\ldots+10) \lg x=5.5
$$
In the parentheses, we have the sum of the terms of an arithmetic progression $S_{n}$ with $a_{1}=1$, $d=1, a_{n}=10, n=10: S_{n}=\frac{a_{1}+a_{n}}{2} n=\frac{1+10}{2... | \sqrt[10]{10} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,366 |
7.192. $\sqrt{3 \log _{2}^{2} x-1-9 \log _{x}^{2} 2}=5$. | Solution.
Domain of definition: $\left\{\begin{array}{l}3 \log _{2}^{4} x-\log _{2}^{2} x-9 \geq 0, \\ 0<x \neq 1\end{array}\right.$
Square both sides of the equation. Then
$\frac{3 \log _{2}^{4} x-\log _{2}^{2} x-9}{\log _{2}^{2} x}=25 \Leftrightarrow 3 \log _{2}^{4} x-26 \log _{2}^{2} x-9=0$.
Solving this equatio... | \frac{1}{8};8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,367 |
7.194. $\log _{x} 2-\log _{4} x+\frac{7}{6}=0$. | ## Solution.
Domain of definition: $0<x \neq 1$
Switch to base 2. We have $\frac{1}{\log _{2} x}-\frac{1}{2} \log _{2} x+\frac{7}{6}=0 \Leftrightarrow$ $\Leftrightarrow 3 \log _{2}^{2} x-7 \log _{2} x-6=0$. Solving this equation as a quadratic in terms of $\log _{2} x$, we find $\left(\log _{2} x\right)_{1}=-\frac{2}... | \frac{1}{\sqrt[3]{4}};8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,369 |
7.195. $\log _{x}(125 x) \cdot \log _{25}^{2} x=1$. | ## Solution.
Domain of definition: $0<x \neq 1$.
Switch to base 5. Then we get $\frac{\log _{5} 125 x}{\log _{5} x} \cdot \frac{\log _{5}^{2} x}{\log _{5}^{2} 25}=1 \Leftrightarrow$
$\Leftrightarrow \log _{5}^{2} x+3 \log _{5} x-4=0$. Solving this equation as a quadratic equation in terms of $\log _{5} x$, we have $\... | \frac{1}{625};5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,370 |
7.196. $3^{\log _{3} x+\log _{3} x^{2}+\log _{3} x^{3}+\ldots+\log _{3} x^{8}}=27 x^{30}$. | Solution.
Domain of definition: $x>0$.
Rewrite the equation as $3^{\log _{3} x+2 \log _{3} x+3 \log _{3} x+\ldots+8 \log _{3} x}=27 x^{30} \Leftrightarrow$ $\Leftrightarrow\left(3^{\log _{3} x}\right)^{(1+2+3+\ldots+8)}=27 x^{30} \Leftrightarrow x^{1+2+3+\ldots+8}=27 x^{30} \Leftrightarrow x^{36}=27 x^{30} \Leftright... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,371 |
7.197. $5^{\frac{x}{\sqrt{x}+2}} \cdot 0.2^{\frac{4}{\sqrt{x}+2}}=125^{x-4} \cdot 0.04^{x-2}$. | ## Solution.
Domain of definition: $x \geq 0$.
From the condition we have
$$
\begin{aligned}
& 5^{\frac{x}{\sqrt{x}+2}} \cdot 5^{-\frac{4}{\sqrt{x}+2}}=5^{3 x-12} \cdot 5^{-2 x+4} \Leftrightarrow 5^{\frac{x}{\sqrt{x}+2}-\frac{4}{\sqrt{x}+2}}=5^{3 x-12-2 x+4} \Leftrightarrow \\
& \Leftrightarrow \frac{x-4}{\sqrt{x}+2... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,372 |
7.198. $\left(3 \cdot\left(3^{\sqrt{x}+3}\right)^{\frac{1}{2 \sqrt{x}}}\right)^{\frac{2}{\sqrt{x}-1}}=\frac{3}{\sqrt[10]{3}}$. | ## Solution.
Domain of definition: $0<x \neq 1$.
From the condition
$$
3^{\frac{3(\sqrt{x}+1)}{2 \sqrt{x}} \cdot \frac{2}{\sqrt{x}-1}}=3^{\frac{9}{10}} \Leftrightarrow \frac{3(\sqrt{x}+1)}{2 \sqrt{x}} \cdot \frac{2}{\sqrt{x}-1}=\frac{9}{10} \Leftrightarrow 3 x-13 \sqrt{x}-10=0
$$
Solving this equation as a quadrati... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,373 |
7.199. $\log _{2} \log _{3}\left(x^{2}-16\right)-\log _{1 / 2} \log _{1 / 3} \frac{1}{x^{2}-16}=2$. | ## Solution.
Domain of definition: $\log _{3}\left(x^{2}-16\right)>0 \Leftrightarrow x^{2}-16>3 \Leftrightarrow x^{2}>19 \Leftrightarrow$ $\Leftrightarrow x \in(-\infty ;-\sqrt{19}) \cup(\sqrt{19} ; \infty)$.
Rewrite the equation as
$\log _{2} \log _{3}\left(x^{2}-16\right)+\log _{2} \log _{3}\left(x^{2}-16\right)=2... | -5;5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,374 |
7.200. $\frac{1+2 \log _{9} 2}{\log _{9} x}-1=2 \log _{x} 3 \cdot \log _{9}(12-x)$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}0<x<12, \\ x \neq 1 .\end{array}\right.$
Switch to base 3. Then we get
$$
\begin{aligned}
& \frac{1+\frac{2 \log _{3} 2}{\log _{3} 9}}{\frac{\log _{3} x}{\log _{3} 9}}-1=\frac{2}{\log _{3} x} \cdot \frac{\log _{3}(12-x)}{\log _{3} 9} \Leftrightarrow \frac{2+... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,375 |
7.201. $3 \lg 2+\lg \left(2^{\sqrt{x-1}-1}-1\right)=\lg \left(0.4 \sqrt{2^{\sqrt{x-1}}}+4\right)+1$. | Solution.
Domain of definition: $\left\{\begin{array}{l}2^{\sqrt{x-1}-1}-1>0, \\ x-1 \geq 0\end{array} \Leftrightarrow x>2\right.$.
Rewrite the equation in the form
$$
\begin{aligned}
& \lg 8+\lg \left(2^{\sqrt{x-1}-1}-1\right)=\lg \left(0.4 \sqrt{2^{\sqrt{x-1}}}+4\right)+\lg 10 \Leftrightarrow \\
& \Leftrightarrow ... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,376 |
7.202. $5 \log _{x / 9} x+\log _{9 / x} x^{3}+8 \log _{9 x^{2}} x^{2}=2$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ x \neq \frac{1}{9}, \\ x \neq \pm \frac{1}{3} .\end{array}\right.$
Switch to base 9. We have
$$
\begin{aligned}
& \frac{5 \log _{9} x}{\log _{9} \frac{x}{9}}+\frac{\log _{9} x^{3}}{\log _{9} \frac{9}{x}}+\frac{8 \log _{9} x^{2}}{\log _{9} 9 x^{2}}=2 ... | \sqrt{3};3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,377 |
7.203. $20 \log _{4 x} \sqrt{x}+7 \log _{16 x} x^{3}-3 \log _{x / 2} x^{2}=0$.
7.203. $20 \log _{4 x} \sqrt{x}+7 \log _{16 x} x^{3}-3 \log _{x / 2} x^{2}=0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ x \neq \frac{1}{4}, \\ x \neq \frac{1}{16}, \\ x \neq 2 .\end{array}\right.$
Transition to base 2:
$$
\begin{aligned}
& \frac{20 \log _{2} \sqrt{x}}{\log _{2} 4 x}+\frac{7 \log _{2} x^{3}}{\log _{2} 16 x}-\frac{3 \log _{2} x^{3}}{\log _{2} \frac{x}{2}}=... | 1;\frac{1}{4\sqrt[5]{8}};4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,378 |
7.204. $\sqrt[4]{|x-3|^{x+1}}=\sqrt[3]{|x-3|^{x-2}}$.
7.204. $\sqrt[4]{|x-3|^{x+1}}=\sqrt[3]{|x-3|^{x-2}}$. | Solution.
Obviously, $x \neq 3$, then $|x-3|>0$. Rewrite the equation as $|x-3|^{\frac{x+1}{4}}=|x-3|^{\frac{x-2}{3}}$
We get two cases:
$$
\begin{aligned}
& \text { 1) }|x-3|=1 \Rightarrow x_{1}=2, x_{2}=4 \\
& \text { 2) }|x-3| \neq 1 \Rightarrow \frac{x+1}{4}=\frac{x-2}{3} \Leftrightarrow 3 x+3=4 x-8, x_{3}=11
\e... | 2;4;11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,379 |
7.205. $|x-3|^{3 x^{2}-10 x+3}=1$.
7.205. $|x-3|^{3 x^{2}-10 x+3}=1$. | ## Solution.
Obviously, $x \neq 3$, hence $|x-3|>0$. Taking the logarithm of both sides of the equation with base 10, we have $\left(3 x^{2}-10 x+3\right) \lg |x-3|=0$, from which $3 x^{2}-10 x+3=0$ or $\lg |x-3|=0$. The roots of the quadratic equation $3 x^{2}-10 x+3=0$ are $x_{1}=\frac{1}{3}$ and $x_{2}=3$. From the... | \frac{1}{3};2;4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,380 |
7.206. $|x-2|^{10 x^{2}-3 x-1}=1$.
7.206. $|x-2|^{10 x^{2}-3 x-1}=1$. | Solution.
Rewrite the equation as $|x-2|^{10 x^{2}-3 x-1}=|x-2|^{0}$. Then we get two cases:
1) $|x-2|=1$, from which $x-2=-1$ or $x-2=1, x_{1}=1, x_{2}=3$;
2) $\left\{\begin{array}{l}0<|x-2| \neq 1, \\ 10 x^{2}-3 x-1=0\end{array} \Leftrightarrow\left\{\begin{array}{l}x \neq 2, \\ x \neq 1, x \neq 3, \\ x_{3}=-\frac{... | -\frac{1}{5};\frac{1}{2};1;3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,381 |
7.207. $\log _{\sqrt{a}} \frac{\sqrt{2 a-x}}{a}-\log _{1 / a} x=0$.
7.207. $\log _{\sqrt{a}} \frac{\sqrt{2 a-x}}{a}-\log _{1 / a} x=0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}0<a \neq 1, \\ 2 a-x \geq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}0<\dot{a} \neq 1, \\ x \leq 2 a .\end{array}\right.\right.$
From the condition we have
$$
\begin{aligned}
& \frac{\log _{a} \frac{\sqrt{2 a-x}}{a}}{\log _{a} \sqrt{a}}-\frac{\log _{a} ... | a | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,382 |
7.208. $2^{x-1}+2^{x-4}+2^{x-2}=6.5+3.25+1.625+\ldots$ (the expression on the right side is an infinite geometric progression). | Solution.
On the right side, we have the sum of the terms of an infinitely decreasing geometric progression $S$, where $b_{1}=6.5 ; q=\frac{3.25}{6.5}=0.5 \Rightarrow S=\frac{b_{1}}{1-q}=\frac{6.5}{1-0.5}=13$.
Rewrite the equation as $\frac{2^{x}}{2}+\frac{2^{x}}{16}+\frac{2^{x}}{4}=13 \Leftrightarrow \frac{13}{16} \... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,383 |
7.209. $49^{1+\sqrt{x-2}}-344 \cdot 7^{\sqrt{x-2}}=-7$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
7.209. $49^{1+\sqrt{x-2}}-344 \cdot 7^{\sqrt{x-2}}=-7$. | Solution.
Domain of definition: $x \geq 2$.
Rewrite the equation as $49 \cdot 7^{2 \sqrt{x-2}}-344 \cdot 7^{\sqrt{x-2}}+7=0$. Solving it as a quadratic equation in terms of $7^{\sqrt{x-2}}$, we get $\left(7^{\sqrt{x-2}}\right)=7^{-2}$ or $\left(7^{\sqrt{x-2}}\right)_{2}=7$, from which $(\sqrt{x-2})_{1}=-2$ (no soluti... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,384 |
7.210. $5^{x-1}+5 \cdot 0.2^{x-2}=26$. | ## Solution.
Let's rewrite the equation as
$$
\frac{5^{x}}{5}+\frac{125}{5^{x}}-26 \doteq 0 \Leftrightarrow 5^{2 x}-130 \cdot 5^{x}+625=0
$$
Solving it as a quadratic equation in terms of $5^{x}$, we get $\left(5^{x}\right)_{1}=5$ or $\left(5^{x}\right)_{2}=5^{3}$, from which $x_{1}=1, x_{2}=3$.
Answer: $1 ; 3$. | 1;3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,385 |
7.211. $\log _{\sqrt{3}} x \cdot \sqrt{\log _{\sqrt{3}} 3-\log _{x} 9}+4=0$. | Solution.
Domain of definition: $\left[\begin{array}{l}x \geq 3, \\ 0<x<1 .\end{array}\right.$
Let's transition to base 3. We get
$$
\begin{aligned}
& 2 \log _{3} x \cdot \sqrt{2-\frac{2}{\log _{3} x}}+4=0 \Leftrightarrow \log _{3} x \cdot \sqrt{\frac{2 \log _{3} x-2}{\log _{3} x}}=-2 \Rightarrow \\
& \Rightarrow\le... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,386 |
7.212. $\frac{\log _{4 \sqrt{x}} 2}{\log _{2 x} 2}+\log _{2 x} 2 \cdot \log _{1 / 2} 2 x=0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ x \neq \frac{1}{16}, \\ x \neq \frac{1}{2}, \\ x \neq 1 .\end{array}\right.$
Transition to base 2. We have
$\frac{\frac{\log _{2} 2}{\log _{2} 4 \sqrt{x}}}{\frac{\log _{2} 2}{\log _{2} 2 x}}+\frac{\log _{2} 2}{\log _{2} 2 x} \cdot \frac{\log _{2} 2 x}{\... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,387 |
7.213. $\left|\log _{\sqrt{3}} x-2\right|-\left|\log _{3} x-2\right|=2$. | ## Solution.
Domain of definition: $x>0$.
Switch to base 3. Then $\left|2 \log _{3} x-2\right|-\left|\log _{3} x-2\right|=2$. By opening the absolute values, we get three cases:
1) $\left\{\begin{array}{l}\log _{3} x<1, \\ -2 \log _{3} x+2+\log _{3} x-2=2\end{array} \Leftrightarrow\left\{\begin{array}{l}\log _{3} x<... | \frac{1}{9};9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,388 |
7.214. $9^{x}+6^{x}=2^{2 x+1}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
7.214. $9^{x}+6^{x}=2^{2 x+1}$. | Solution.
Rewrite the equation as $3^{2 x}+2^{x} \cdot 3^{x}-2 \cdot 2^{2 x}=0$ and divide it by $2^{2 x} \neq 0$. Then $\left(\frac{3}{2}\right)^{2 x}+\left(\frac{3}{2}\right)^{x}-2=0 \Rightarrow\left(\left(\frac{3}{2}\right)^{x}\right)=-2$ (no solutions) or $\left(\left(\frac{3}{2}\right)^{x}\right)_{2}=1 \Rightarro... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,389 |
7.215. $2^{x+\sqrt{x^{2}-4}}-5 \cdot(\sqrt{2})^{x-2+\sqrt{x^{2}-4}}-6=0$. | ## Solution.
Domain of definition: $x^{2}-4 \geq 0 \Leftrightarrow x \in(-\infty ;-2] \cup[2 ; \infty)$.
Let's write the equation in the form $2^{x+\sqrt{x^{2}-4}}-\frac{5}{2} \cdot 2^{\frac{x+\sqrt{x^{2}-4}}{2}}-6=0$. Solving it as a quadratic equation in terms of $2^{\frac{x+\sqrt{x^{2}-4}}{2}}$, we get $2^{\frac{x... | \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,390 |
7.216. $27^{x}-13 \cdot 9^{x}+13 \cdot 3^{x+1}-27=0$. | ## Solution.
We have
$$
\begin{aligned}
& 3^{3 x}-13 \cdot 3^{2 x}+39 \cdot 3^{x}-27=0 \Leftrightarrow\left(3^{3 x}-27\right)-13 \cdot 3^{x}\left(3^{x}-3\right)=0 \Leftrightarrow \\
& \Leftrightarrow\left(3^{x}-3\right)\left(3^{2 x}+3 \cdot 3^{x}+9\right)-13 \cdot 3^{x}\left(3^{x}-3\right)=0 \Leftrightarrow \\
& \Lef... | 0;1;2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,391 |
7.218. $\left(\frac{3}{5}\right)^{2 \log _{9}(x+1)} \cdot\left(\frac{125}{27}\right)^{\log _{27}(x-1)}=\frac{\log _{5} 27}{\log _{5} 243}$. | Solution.
Domain of definition: $x>1$.
From the condition we have
$\left(\frac{3}{5}\right)^{\log _{3}(x+1)} \cdot\left(\frac{3}{5}\right)^{\log _{3}(x-1)}=\frac{3}{5} \Leftrightarrow\left(\frac{3}{5}\right)^{\log _{3}(x+1)+\log _{3}(x-1)}=\frac{3}{5} \Rightarrow$
$\Rightarrow \log _{3}(x+1)+\log _{3}(x-1)=1 \Right... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,393 |
7.219. $5^{1+x^{3}}-5^{1-x^{3}}=24$.
7.219. $5^{1+x^{3}}-5^{1-x^{3}}=24$. | Solution.
We have $5 \cdot 5^{x^{3}}-\frac{5}{5^{x^{3}}}-24=0 \Leftrightarrow 5 \cdot\left(5^{x^{3}}\right)^{2}-24 \cdot 5^{x^{3}}-5=0$. Solving this equation as a quadratic in terms of $5^{x^{3}}$, we get $5^{x^{3}}=-\frac{1}{5}$ (no solutions), or $5^{x^{3}}=5 \Rightarrow x^{3}=1, x=1$.
Answer: 1 . | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,394 |
7.220. $3^{2 x+4}+45 \cdot 6^{x}-9 \cdot 2^{2 x+2}=0$.
7.220. $3^{2 x+4}+45 \cdot 6^{x}-9 \cdot 2^{2 x+2}=0$.
(No change needed as the text is already in English and contains only a mathematical equation.) | Solution.
Rewrite the equation as $81 \cdot 3^{2 x}+45 \cdot 3^{x} \cdot 2^{x}-36 \cdot 2^{x}=0$. Dividing it by $9 \cdot 2^{2 x}$, we get $9 \cdot\left(\frac{3}{2}\right)^{2 x}+5 \cdot\left(\frac{3}{2}\right)^{x}-4=0 \Rightarrow\left(\frac{3}{2}\right)^{x}=-1$ (no solutions), or $\left(\frac{3}{2}\right)^{x}=\left(\f... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,395 |
7.221. $4^{\lg x+1}-6^{\lg x}-2 \cdot 3^{\lg x^{2}+2}=0$.
7.221. $4^{\lg x+1}-6^{\lg x}-2 \cdot 3^{\lg x^{2}+2}=0$. | ## Solution.
Domain of definition: $x>0$.
From the condition, we have $4 \cdot 2^{2 \lg x}-2^{\lg x} \cdot 3^{\lg x}-18 \cdot 3^{2 \lg x}=0$. Dividing it by $3^{2 \lg x}$, we get $4 \cdot\left(\frac{2}{3}\right)^{2 \lg x}-\left(\frac{2}{3}\right)^{\lg x}-18=0 \Rightarrow\left(\frac{2}{3}\right)^{\lg x}=-2$ (no soluti... | 0.01 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,396 |
7.222. $3 \cdot 16^{x}+2 \cdot 81^{x}=5 \cdot 36^{x}$. | ## Solution.
We have $3 \cdot 4^{2 x}+2 \cdot 9^{2 x}-5 \cdot 4^{x} \cdot 9^{x}=0 \Rightarrow 3 \cdot\left(\frac{4}{9}\right)^{2 x}-5 \cdot\left(\frac{4}{9}\right)^{x}+2=0 \Rightarrow$ $\Rightarrow\left(\frac{4}{9}\right)^{x}=\frac{2}{3}$ or $\left(\frac{4}{9}\right)^{x}=1$, from which $x_{1}=\frac{1}{2}, x_{2}=0$.
A... | 0;\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,397 |
7.223. $\log _{5}\left(2^{1.5 x-2.5}+2^{1.5 x-0.5}-0.01 \cdot 5^{3 x+1}\right)=3 x-1$. | ## Solution.
Domain of definition: $2^{1.5 x-2.5}+2^{1.5 x-0.5}-0.01 \cdot 5^{3 x+1}>0$. By the definition of the logarithm, we get
$$
\begin{aligned}
& 2^{1.5 x-2.5}+2^{1.5 x-0.5}-0.01 \cdot 5^{3 x+1}=5^{3 x-1} \Leftrightarrow \\
& \Leftrightarrow \frac{2^{1.5 x}}{2^{2.5}}+\frac{2^{1.5 x}}{2^{0.5}}-0.05 \cdot 5^{3 x... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,398 |
7.224. $\frac{8^{x}+2^{x}}{4^{x}-2}=5$ | Solution.
Domain of definition: $x \neq \frac{1}{2}$.
Rewrite the equation as $2^{3 x}-5 \cdot 2^{2 x}+2^{x}+10=0$. Let $2^{x}=y$. Then the equation becomes $y^{3}-5 y^{2}+y+10=0$. Divide the left side of the equation by $y-2$:
-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,399 |
7.225. $\log _{3 x+7}(5 x+3)+\log _{5 x+3}(3 x+7)=2$. | Solution.
Domain of definition: $\left\{\begin{array}{l}0-\frac{3}{5}, x \neq-\frac{2}{5}\right.$.
Multiplying the equation by $\log _{3 x+7}(5 x+3) \neq 0$, we get
$$
\begin{aligned}
& \log _{3 x+7}^{2}(5 x+3)-2 \log _{3 x+7}(5 x+3)+1=0 \Leftrightarrow\left(\log _{3 x+7}(5 x+3)-1\right)^{2}=0 \Leftrightarrow \\
& \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,400 |
7.226. $2.5^{\log _{3} x}+0.4^{\log _{3} x}=2.9$. | ## Solution.
Domain of definition: $x>0$.
Rewrite the equation as $\left(\frac{5}{2}\right)^{\log _{3} x}+\left(\frac{2}{5}\right)^{\log _{3} x}-2.9=0$. Multiplying the equation by $\left(\frac{5}{2}\right)^{\log _{3} x}$, we get $\left(\frac{5}{2}\right)^{2 \log _{3} x}-2.9 \cdot\left(\frac{5}{2}\right)^{\log _{3} x... | \frac{1}{3};3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,401 |
7.227. $(\lg (x+20)-\lg x) \log _{x} 0.1=-1$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x+20>0, \\ 0<x \neq 1\end{array}\right.$ or $0<x \neq 1$.
Switch to base 10. We have $(\lg (x+20)-\lg x)\left(-\frac{1}{\lg x}\right)=-1 \Leftrightarrow$ $\Leftrightarrow \lg (x+20)-\lg x=\lg x \Leftrightarrow \lg (x+20)=2 \lg x \Leftrightarrow \lg (x+20)=\lg x^... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,402 |
7.228. $5^{\lg x}=50-x^{\lg 5}$. | ## Solution.
Domain of definition: $0<x \neq 1$.
Rewrite the equation as $5^{\lg x}=50-5^{\lg x}, 2 \cdot 5^{\lg x}=50, 5^{\lg x}=25$, from which $\lg x=2, x=10^{2}=100$.
Answer: 100. | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,403 |
7.229. $27 \cdot 2^{-3 x}+9 \cdot 2^{x}-2^{3 x}-27 \cdot 2^{-x}=8$. | Solution.
Transform the equation:
$$
\begin{aligned}
& 27+9 \cdot 2^{4 x}-2^{6 x}-27 \cdot 2^{2 x}=8 \cdot 2^{3 x} \Leftrightarrow \\
& \Leftrightarrow 2^{6 x}-9 \cdot 2^{4 x}+8 \cdot 2^{3 x}+27 \cdot 2^{2 x}-27=0 \Leftrightarrow \\
& \Leftrightarrow 2^{6 x}-2^{4 x}-8 \cdot 2^{4 x}+8 \cdot 2^{3 x}+27 \cdot 2^{x}-27=0... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,404 |
7.230. $\log _{x+1}(x-0.5)=\log _{x-0.5}(x+1)$. | Solution.
Domain of definition: $\left\{\begin{array}{l}0<x+1 \neq 1, \\ 0<x-0.5 \neq 1\end{array}\right.$ or $0.5<x \neq 1.5$.
Multiplying both sides of the equation by $\log _{x+1}(x-0.5) \neq 0$, we get
$$
\begin{aligned}
& \log _{x+1}^{2}(x-0.5)=1 \Rightarrow \log _{x+1}(x-0.5)=-1 \Rightarrow \\
& \Rightarrow x-... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,405 |
7.231. $\log _{4} \log _{2} x+\log _{2} \log _{4} x=2$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\log _{2} x>0, \\ \log _{4} x>0,\end{array} \Leftrightarrow x>1\right.$.
Switch to base 2. We have
$$
\begin{aligned}
& \frac{1}{2} \log _{2} \log _{2} x+\log _{2}\left(\frac{1}{2} \log _{2} x\right)=2 \Leftrightarrow \\
& \Leftrightarrow \log _{2} \log _{2} x+... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,406 |
7.232. $\log _{x^{2}} 16+\log _{2 x} 64=3$. | Solution.
Domain of definition: $\left\{\begin{array}{l}0<x \neq \frac{1}{2}, \\ x \neq 1 .\end{array}\right.$
Transition to base 2. Then
$$
\begin{aligned}
& \frac{\log _{2} 16}{\log _{2} x^{2}}+\frac{\log _{2} 64}{\log _{2} 2 x}=3 \Leftrightarrow \frac{4}{2 \log _{2} x}+\frac{6}{1+\log _{2} x}-3=0 \Leftrightarrow ... | 0.5\sqrt[3]{4};4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,407 |
7.233. $\left(3 \log _{a} x-2\right) \log _{x}^{2} a=\log _{\sqrt{a}} x-3 \quad(a>0, a \neq 1)$. | Solution.
Domain of definition: $\left\{\begin{array}{l}0<a \neq 1, \\ 0<x \neq 1 .\end{array}\right.$
Let's transition to the base $a$. We get
$$
\frac{3 \log _{a} x-2}{\log _{a}^{2} x}=2 \log _{a} x-3 \Leftrightarrow 2 \log _{a}^{3} x-3 \log _{a}^{2} x-3 \log _{a} x+2=0
$$
since $\log _{a} x \neq 0$. Further, we ... | \frac{1}{};\sqrt{};^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,408 |
7.234. $\frac{10 x^{2 \lg x}}{x^{3}}=\frac{x^{3 \lg x}}{10}$. | ## Solution.
Domain of definition: $0<x \neq 1$.
From the condition, we have $\frac{x^{2 \lg ^{2} x}}{x^{3} \cdot x^{3 \lg x}}=\frac{1}{100} \Leftrightarrow x^{2 \lg ^{2} x-3 \lg x-3}=10^{-2}$. Taking the logarithm of both sides of this equation with base 10, we get
$$
\begin{aligned}
& \lg x^{2 \lg ^{2} x-3 \lg x-3... | \frac{1}{10},\sqrt{10},100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,409 |
7.235. $x \log _{x+1} 5 \cdot \log _{\sqrt[3]{1 / 5}}(x+1)=\frac{x-4}{x}$. | ## Solution.
$$
\text { Domain: }\left\{\begin{array}{l}
0<x+1 \neq 1, \\
x \neq 0
\end{array} \Leftrightarrow-1<x \neq 0\right.
$$
Switch to base 5. We have $\frac{x}{\log _{5}(x+1)} \cdot(-3) \log _{5}(x+1)=\frac{x-4}{x}$,
$-3 x=\frac{x-4}{x}$ when $\log _{5}(x+1) \neq 0$. From here, $3 x^{2}+x-4=0, x_{1}=-\frac{4... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,410 |
7.237. $4 \log _{4}^{2}(-x)+2 \log _{4}\left(x^{2}\right)=-1$.
7.237. $4 \log _{4}^{2}(-x)+2 \log _{4}\left(x^{2}\right)=-1$. | Solution.
Domain of definition: $\left\{\begin{array}{l}-x>0, \\ x^{2}>0\end{array} \Leftrightarrow x<0\right.$.
Since by the domain of definition $x<0$, we have
$4 \log _{4}^{2}(-x)+4 \log _{4}(-x)+1=0 \Leftrightarrow\left(2 \log _{4}(-x)+1\right)^{2}=0 \Leftrightarrow$
$\Leftrightarrow 2 \log _{4}(-x)=-1, \log _{... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,412 |
7.240. $\lg ^{4}(x-1)^{2}+\lg ^{2}(x-1)^{3}=25$. | ## Solution.
Domain of definition: $x>1$.
From the condition, we have $16 \lg ^{4}(x-1)+9 \lg ^{2}(x-1)-25=0$. Solving this equation as a biquadratic equation in terms of $\lg (x-1)$, we get $\lg ^{2}(x-1)=1 \Rightarrow$ $\Rightarrow \lg (x-1)=-1$ or $\lg (x-1)=1$, from which $x_{1}=1.1, x_{2}=11$.
Answer: 1,$1 ; 11... | 1.1;11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,415 |
7.241. $\frac{\log _{2}\left(x^{3}+3 x^{2}+2 x-1\right)}{\log _{2}\left(x^{3}+2 x^{2}-3 x+5\right)}=\log _{2 x} x+\log _{2 x} 2$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x^{3}+3 x^{2}+2 x-1>0, \\ 0<x^{3}+2 x^{2}-3 x+5 \neq 1, \\ 0<x \neq \frac{1}{2} .\end{array}\right.$
By the formula for changing the base we have
$$
\begin{aligned}
& \log _{x^{3}+2 x^{2}-3 x+5}\left(x^{3}+3 x^{2}+2 x-1\right)=\log _{2 x} 2 x \Leftrightarrow... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,416 |
7.242. $\left(16 \cdot 5^{2 x-1}-2 \cdot 5^{x-1}-0.048\right) \lg \left(x^{3}+2 x+1\right)=0$. | Solution.
Domain: $x^{3}+2 x+1>0$.
From the condition $16 \cdot 5^{2 x-1}-2^{x-1}-0.048=0$ or $\lg \left(x^{3}+2 x+1\right)=0$. Rewrite the first equation as
$\frac{16}{5} \cdot 5^{2 x}-\frac{2}{5} \cdot 5^{x}-0.048=0 \Leftrightarrow 16 \cdot 5^{2 x}-2 \cdot 5^{x}-0.24=0$.
Solving this equation as a quadratic in te... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,417 |
7.243. $5^{x} \cdot \sqrt[x]{8^{x-1}}=500$. | Solution.
Rewrite the equation as $5^{x} \cdot 8^{\frac{x-1}{x}}=500 \Leftrightarrow \frac{5^{x} \cdot 8}{8^{1 / x}}=500 \Leftrightarrow$ $\Leftrightarrow \frac{5^{x} \cdot 2}{8^{1 / x}}=125 \Leftrightarrow 5^{x-3}=2^{3 / x-1} \Rightarrow\left\{\begin{array}{l}x-3=0, \\ \frac{3}{x}-1=0,\end{array}\right.$ from which $x... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,418 |
7.244. $3 \log _{2}^{2} \sin x+\log _{2}(1-\cos 2 x)=2$. | Solution.
Domain of definition: $0<\sin x<1$.
Since $1-\cos 2 x=2 \sin ^{2} x$, we have $3 \log _{2}^{2} \sin x+\log _{2} 2 \sin ^{2} x-2=0 \Leftrightarrow 3 \log _{2}^{2} \sin x+2 \log _{2} \sin x-1=0$.
Solving this equation as a quadratic in $\log _{2} \sin x$, we get $\log _{2} \sin x=\frac{1}{3}$ or $\log _{2} \... | (-1)^{n}\frac{\pi}{6}+\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,419 |
7.245. $\log _{1+x}\left(2 x^{3}+2 x^{2}-3 x+1\right)=3$. | Solution.
Domain of definition: $\left\{\begin{array}{l}2 x^{3}+2 x^{2}-3 x+1>0, \\ -1<x \neq 0 .\end{array}\right.$
We have
$$
\begin{aligned}
& 2 x^{3}+2 x^{2}-3 x+1=(1+x)^{3} \Leftrightarrow 2 x^{3}+2 x^{2}-3 x+1=1+3 x+3 x^{2}+x^{3} \Leftrightarrow \\
& \Leftrightarrow x^{3}-x^{2}-6 x=0 \Leftrightarrow x\left(x^{... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,420 |
7.246. $\log _{2} \sqrt[3]{x}+\sqrt[3]{\log _{2} x}=\frac{4}{3}$. | ## Solution.
Domain of definition: $x>0$.
## From the condition we have
$$
\frac{1}{3} \log _{2} x+\sqrt[3]{\log _{2} x}=\frac{4}{3} \Leftrightarrow \log _{2} x+3 \sqrt[3]{\log _{2} x}-4=0
$$
Let $\sqrt[3]{\log _{2} x}=y$. The equation in terms of $y$ becomes
$$
\begin{aligned}
& y^{3}+3 y-4=0 \Leftrightarrow\left... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,421 |
7.247. $\sqrt{\log _{5} x}+\sqrt[3]{\log _{5} x}=2$. | Solution.
Domain of definition: $\log _{5} x \geq 0$ or $x \geq 1$.
Rewrite the equation as $\sqrt[6]{\left(\log _{5} x\right)^{3}}+\sqrt[6]{\left(\log _{5} x\right)^{2}}-2=0$. Let $\sqrt[6]{\log _{5} x}=y$. The equation in terms of $y$ becomes
$$
y^{3}+y^{2}-2=0 \Leftrightarrow\left(y^{3}-1\right)+\left(y^{2}-1\rig... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,422 |
7.248. $\log _{2} x \cdot \log _{3} x=\log _{3}\left(x^{3}\right)+\log _{2}\left(x^{2}\right)-6$. | ## Solution.
Domain of definition: $x>0$.
Let's switch to base 2. We have
$$
\begin{aligned}
& \frac{\log _{2} x \cdot \log _{2} x}{\log _{2} 3}=\frac{3 \log _{2} x}{\log _{2} 3}+2 \log _{2} x-6 \Leftrightarrow \\
& \Leftrightarrow \log _{2}^{2} x-\left(3+2 \log _{2} 3\right) \log _{2} x+6 \log _{2} 3=0
\end{aligned... | 8;9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,423 |
7.249. $3 \cdot 4^{(x-2)}+27=a+a \cdot 4^{(x-2)}$. For what values of $a$ does the equation have a solution? | ## Solution.
Rewrite the equation as
$$
\begin{aligned}
& 3 \cdot 4^{(x-2)}-a \cdot 4^{(x-2)}=a-27 \Leftrightarrow (3-a) \cdot 4^{(x-2)}=a-27 \Rightarrow \\
& \Rightarrow 4^{(x-2)}=\frac{a-27}{3-a}, \text{ where } \frac{a-27}{3-a}>0 .
\end{aligned}
$$
Taking the logarithm of both sides of this equation with base 4, ... | \in(3;27) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,424 |
7.250. $\log _{a} x+\log _{\sqrt{a}} x+\log _{\sqrt[3]{a^{2}}} x=27$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ 0<a \neq 1 .\end{array}\right.$
Switch to base $a$. We have $\log _{a} x+2 \log _{a} x+\frac{3}{2} \log _{a} x=27 \Leftrightarrow$ $\log _{a} x=6$, from which $x=a^{6}$.
Answer: $a^{6}$, where $0<a \neq 1$. | ^6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,425 |
7.251. $x^{2-\lg ^{2} x-\lg x^{2}}-\frac{1}{x}=0$.
7.251. $x^{2-\log ^{2} x-\log x^{2}}-\frac{1}{x}=0$. | ## Solution.
Domain of definition: $x>0$.
Let's write the equation in the form $x^{2-\lg ^{2} x-2 \lg x}=x^{-1}$. Taking the logarithm of both sides of the equation with base 10, we get
$$
\begin{aligned}
& \lg x^{2-\lg ^{2} x-2 \lg x}=\lg x^{-1} \Leftrightarrow\left(2-\lg ^{2} x-2 \lg x\right) \lg x=-\lg x \Leftrig... | 0.001;1;10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,426 |
7.252. $\frac{2}{15}\left(16^{\log _{9} x+1}-16^{\log _{3} \sqrt{x}}\right)+16^{\log _{3} x}-\log _{\sqrt{5}} 5 \sqrt{5}=0$. | ## Solution.
Domain of definition: $x>0$.
Rewrite the equation as
$$
\begin{aligned}
& \frac{2}{15}\left(16 \cdot 16^{\frac{1}{2} \log _{3} x}-16^{\frac{1}{2} \log _{3} x}\right)+16^{\log _{3} x}-3=0 \Leftrightarrow \\
& \Leftrightarrow 16^{\log _{3} x}+2 \cdot 16^{\frac{\log _{3} x}{2}}-3=0
\end{aligned}
$$
Solvin... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,427 |
7.253. $\log _{a} \sqrt{4+x}+3 \log _{a^{2}}(4-x)-\log _{a^{4}}\left(16-x^{2}\right)^{2}=2$. For which values of $a$ does the equation have a solution? | Solution.
Domain of definition: $\left\{\begin{array}{l}-4<x<4, \\ 0<a \neq 1 .\end{array}\right.$
Transition to base $a$. We have
$$
\begin{aligned}
& \frac{1}{2} \log _{a}(4+x)+\frac{3}{2} \log _{a}(4-x)-\frac{1}{2} \log _{a}(4-x)-\frac{1}{2} \log _{a}(4+x)=2 \Leftrightarrow \\
& \Leftrightarrow \log _{a}(4-x)=2 \... | 4-^{2},where\in(0;1)\cup(1;2\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,428 |
7.254. $\log _{2} \sqrt[3]{4}+\log _{8}\left(9^{x+1}-1\right)=1+\log _{8}\left(3^{x+1}+1\right)$. | ## Solution.
Domain of definition: $9^{x+1}-1>0 \Leftrightarrow x>-1$.
Since $\log _{2} \sqrt[3]{4}=\frac{2}{3}$, we have
$$
\begin{aligned}
& \log _{8}\left(9 \cdot 3^{2 x}-1\right)-\log _{8}\left(3 \cdot 3^{x}+1\right)=\frac{1}{3} \Leftrightarrow \log _{8} \frac{9 \cdot 3^{2 x}-1}{3 \cdot 3^{x}+1}=\frac{1}{3} \Lef... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,429 |
7.255. $25^{\log _{4} x}-5^{\log _{16} x^{2}+1}=\log _{\sqrt{3}} 9 \sqrt{3}-25^{\log _{16} x}$. | ## Solution.
Domain of definition: $x>0$.
Rewrite the equation as
$5^{\log _{2} x}-5 \cdot 5^{\frac{1}{2} \log _{2} x}=5-5^{\frac{1}{2} \log _{2} x} \Leftrightarrow 5^{\log _{2} x}-4 \cdot 5^{\frac{1}{2} \log _{2} x}-5=0$.
Solving this equation as a quadratic equation in terms of $5^{\frac{1}{\log _{2} x}}$, we get... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,430 |
7.256. $\left(1+\frac{x}{2}\right) \log _{2} 3-\log _{2}\left(3^{x}-13\right)=2$. | Solution.
Domain: $3^{x}-13>0 \Leftrightarrow x>\log _{3} 13$.
From the condition we have
$\log _{2} 3^{1+\frac{x}{2}}-\log _{2}\left(3^{x}-13\right)=2 \Leftrightarrow \log _{2} \frac{3^{1+\frac{x}{2}}}{3^{x}-13}=2 \Leftrightarrow \frac{3^{1+\frac{x}{2}}}{3^{x}-13}=4 \Leftrightarrow$
$\Leftrightarrow 4 \cdot 3^{x}-... | 4\log_{3}2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,431 |
7.257. $\log _{2} 3 \cdot \log _{3} 4 \cdot \log _{4} 5 \cdot \ldots \cdot \log _{n}(n+1)=10(n \in N)$. | Solution.
Let's switch to base 2.
$\log _{2} 3 \cdot \frac{\log _{2} 4}{\log _{2} 3} \cdot \frac{\log _{2} 5}{\log _{2} 4} \cdot \ldots \cdot \frac{\log _{2}(n+1)}{\log _{2} n}=10 \Leftrightarrow \log _{2}(n+1)=10 \Leftrightarrow$
$\Leftrightarrow n+1=2^{10} \Leftrightarrow n=1024-1=1023$.
Answer: 1023. | 1023 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,432 |
7.258. $2^{\frac{a+3}{a+2}} \cdot 32^{\frac{1}{x(a+2)}}=4^{\frac{1}{x}}$ (consider for all real values of $a$ ). | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ a \neq-2 .\end{array}\right.$
From the condition we have
$$
2^{\frac{a+3}{a+2}} \cdot 2^{\frac{5}{x(a+2)}}=2^{\frac{2}{x}} \Leftrightarrow 2^{\frac{a+3}{a+2}+\frac{5}{x(a+2)}}=2^{\frac{2}{x}} \Leftrightarrow \frac{a+3}{a+2}+\frac{5}{x(a+2)}=\fra... | \frac{2-1}{+3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,433 |
7.259. $\left\{\begin{array}{l}2-\log _{2} y=2 \log _{2}(x+y) \\ \log _{2}(x+y)+\log _{2}\left(x^{2}-x y+y^{2}\right)=1 .\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
2 - \log _{2} y = 2 \log _{2}(x + y) \\
\log _{2}(x + y) + \log _{2}\left(x^{2} - x y + y^{2}\right) = 1 .
\end{array}\righ... | Solution.
Domain of definition: $\left\{\begin{array}{l}x+y>0, \\ y>0, \\ x^{2}-x y+y^{2}>0 .\end{array}\right.$
Write the system of equations as
$$
\begin{aligned}
& \left\{\begin{array}{l}
\log _{2} 4-\log _{2} y=\log _{2}(x+y)^{2}, \\
\log _{2}(x+y)+\log _{2}\left(x^{2}-x y+y^{2}\right)=1 .
\end{array} \Leftright... | (1;1),(\frac{\sqrt[3]{6}}{3};\frac{2\sqrt[3]{6}}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,434 |
7.260. $\left\{\begin{array}{l}3 \cdot\left(\frac{2}{3}\right)^{2 x-y}+7 \cdot\left(\frac{2}{3}\right)^{\frac{2 x-y}{2}}-6=0, \\ \lg (3 x-y)+\lg (y+x)-4 \lg 2=0 .\end{array}\right.$ | Solution.
Domain of Definition (DOD): $\left\{\begin{array}{l}3 x-y>0, \\ y+x>0\end{array}\right.$
Solve the first equation of the system as a quadratic equation with respect to $\left(\frac{2}{3}\right)^{\frac{2 x-y}{2}}$. We have
$\left(\frac{2}{3}\right)^{\frac{2 x-y}{2}}=-3$ (no solutions),
$\left(\frac{2}{3}\r... | (2;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,435 |
7.261. $\left\{\begin{array}{l}\left(0,48^{x^{2}+2}\right)^{2 x-y}=1, \\ \lg (x+y)-1=\lg 6-\lg (x+2 y) .\end{array}\right.$
7.261. $\left\{\begin{array}{l}\left(0.48^{x^{2}+2}\right)^{2 x-y}=1, \\ \log (x+y)-1=\log 6-\log (x+2 y) .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x+y>0, \\ x+2 y>0 .\end{array}\right.$
Rewrite the first equation of the system as
$0.48^{\left(x^{2}+2\right)(2 x-y)}=0.48^{0} \Leftrightarrow\left(x^{2}+2\right)(2 x-y)=0 \Leftrightarrow 2 x-y=0, y=2 x$.
From the second equation of the system we have
$\lg \... | (2;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,436 |
7.262. $\left\{\begin{array}{l}\log _{2}(x-y)=5-\log _{2}(x+y) \\ \frac{\lg x-\lg 4}{\lg y-\lg 3}=-1 .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x-y>0, \\ x+y>0, \\ x>0, \\ y>0, \\ y \neq 3 .\end{array}\right.$
From the condition we have
$$
\left\{\begin{array} { l }
{ \operatorname { log } _ { 2 } ( x - y ) = \operatorname { log } _ { 2 } \frac { 3 2 } { x + y } , } \\
{ \operatorname { lg } \frac { x... | (6;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,437 |
7.263. $\left\{\begin{array}{l}4^{\frac{x}{y}+\frac{y}{x}}=32, \\ \log _{3}(x-y)=1-\log _{3}(x+y)\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
4^{\frac{x}{y}+\frac{y}{x}}=32, \\
\log _{3}(x-y)=1-\log _{3}(x+y)
\end{array}\right.
\] | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0, \\ x-y>0, \\ x+y>0 .\end{array}\right.$
Rewrite the system of equations as
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 2 ^ { \frac { 2 x } { y } + \frac { 2 y } { x } } = 2 ^ { 5 } , } \\
{ \operatorname { log } _ { 3 } ( x - y ... | (2;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,438 |
7.264. $\left\{\begin{array}{l}y^{5 x^{2}-51 x+10}=1, \\ x y=15\end{array}\right.$ | ## Solution.
Domain of definition: $y>0$.
Rewrite the first equation of the system as $y^{5 x^{2}-51 x+10}=y^{0} \Leftrightarrow$ $\Leftrightarrow 5 x^{2}-51 x+10=0$ for $0<y \neq 1$. This system of equations is equivalent to two systems:
1) $\left\{\begin{array}{l}y=1, \\ x y=15\end{array} \Leftrightarrow\left\{\be... | (15,1),(10,1.5),(0.2,75) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,439 |
7.265. $\left\{\begin{array}{l}\log _{x} y=2, \\ \log _{x+1}(y+23)=3 .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}y>0, \\ 0<x \neq 1 .\end{array}\right.$
The system of equations is equivalent to the following:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ y = x ^ { 2 } , } \\
{ y + 23 = ( x + 1 ) ^ { 3 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y=x^{2} \\... | (2;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,440 |
7.266. $\left\{\begin{array}{l}\left(x^{2}+y\right) 2^{y-x^{2}}=1 \\ 9\left(x^{2}+y\right)=6^{x^{2}-y}\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
\left(x^{2}+y\right) 2^{y-x^{2}}=1 \\
9\left(x^{2}+y\right)=6^{x^{2}-y}
\end{array}\right.
\] | ## Solution.
From the first equation, we have $x^{2}+y=2^{x^{2}-y}$. Substituting this value into the second equation of the system, we get $9 \cdot 2^{x^{2}-y}=6^{x^{2}-y} \Leftrightarrow$ $\Leftrightarrow 3^{2}=3^{x^{2}-y} \Leftrightarrow x^{2}-y=2$. Then $x^{2}+y=2^{x^{2}-y}=2^{2}=4$ and the original system of equa... | (\sqrt{3};1),(-\sqrt{3};1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,441 |
7.267. $\left\{\begin{array}{l}y-\log _{3} x=1 \\ x^{y}=3^{12}\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
y - \log_{3} x = 1 \\
x^{y} = 3^{12}
\end{array}\right.
\] | ## Solution.
Domain of definition: $0<x \neq 1$.
By logarithmizing the second equation with base 3, we get
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ y - \log _ { 3 } x = 1 , } \\
{ \log _ { 3 } x ^ { y } = \log _ { 3 } 3 ^ { 12 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y=1+\log _{3} x, \\
y \l... | (\frac{1}{81};-3),(27;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,442 |
7.268. $\left\{\begin{array}{l}9^{\sqrt[4]{x y^{2}}}-27 \cdot 3^{\sqrt{y}}=0, \\ \frac{1}{4} \lg x+\frac{1}{2} \lg y=\lg (4-\sqrt[4]{x}) .\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
9^{\sqrt[4]{x y^{2}}}-27 \cdot 3^{\sqrt{y}}=0, \\
\frac{1}{4} \lg x+\frac{1}{2} \lg y=\lg (4-\sqrt[4]{x}) ... | Solution.
Domain of definition: $\left\{\begin{array}{l}00 .\end{array}\right.$
Let's write the system of equations as
$\left\{\begin{array}{l}3^{2 \sqrt[4]{x} \sqrt{y}}=3^{3+\sqrt{y}} \\ \lg \sqrt[4]{x}+\lg \sqrt{y}=\lg (4-\sqrt[4]{x})\end{array} \Leftrightarrow\left\{\begin{array}{l}2 \sqrt[4]{x} \cdot \sqrt{y}=3+... | (1;9),(16;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,443 |
7.269. $\left\{\begin{array}{l}3^{-x} \cdot 2^{y}=1152 \\ \log _{\sqrt{5}}(x+y)=2 .\end{array}\right.$ | ## Solution.
Domain of definition: $x+y>0$.
Let's write the system of equations as
$$
\left\{\begin{array} { l }
{ 3 ^ { - x } \cdot 2 ^ { y } = 1152 } \\
{ x + y = 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
3^{-x} \cdot 2^{y}=1152 \\
x=5-y
\end{array}\right.\right.
$$
Since $x=5-y$, then
$$
3^{y-5} ... | (-2;7) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,444 |
7.270. $\left\{\begin{array}{l}\lg \left(x^{2}+y^{2}\right)=1+\lg 8, \\ \lg (x+y)-\lg (x-y)=\lg 3 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x+y>0, \\ x-y>0 .\end{array}\right.$
From the condition we have $\left\{\begin{array}{l}\lg \left(x^{2}+y^{2}\right)=\lg 80, \\ \lg \frac{x+y}{x-y}=\lg 3\end{array} \Leftrightarrow\left\{\begin{array}{l}x^{2}+y^{2}=80, \\ \frac{x+y}{x-y}=3 .\end{array}\right.... | (8;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,445 |
7.271. $\left\{\begin{array}{l}3^{x} \cdot 2^{y}=972 \\ \log _{\sqrt{3}}(x-y)=2 .\end{array}\right.$ | ## Solution.
Domain of definition: $x-y>0$.
From the second equation of the system, we find $x-y=3, x=y+3$. Substituting this value of $x$ into the first equation, we get $3^{y+3} \cdot 2^{y}=972$, $27 \cdot 3^{y} \cdot 2^{y}=972,6^{y}=6^{2}$, from which $y=2$. Then $x=5$.
Answer: $(5 ; 2)$. | (5;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,446 |
7.272. $\left\{\begin{array}{l}3^{1+2 \log _{3}(y-x)}=48, \\ 2 \log _{5}(2 y-x-12)-\log _{5}(y-x)=\log _{5}(y+x) .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}y-x>0 \\ y+x>0 \\ 2 y-x-12>0 .\end{array}\right.$
Rewrite the system of equations as
$$
\begin{aligned}
& \left\{\begin{array}{l}
3 \cdot 3^{\log _{3}(y-x)^{2}}=48, \\
\log _{5}(2 y-x-12)^{2}-\log _{5}(y-x)=\log _{5}(y+x)
\end{array} \Leftrightarrow\right. \\
&... | (16;20) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,447 |
7.273. $\log _{9}\left(x^{3}+y^{3}\right)=\log _{3}\left(x^{2}-y^{2}\right)=\log _{3}(x+y)$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x+y>0, \\ x-y>0 .\end{array}\right.$
Rewrite the given double equality as a system of equations $\left\{\begin{array}{l}\log _{9}\left(x^{3}+y^{3}\right)=\log _{3}(x+y) \\ \log _{3}\left(x^{2}-y^{2}\right)=\log _{3}(x+y)\end{array}\right.$ In the first equati... | (1;0),(2;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,448 |
7.274. $\left\{\begin{array}{l}\left(\log _{a} x+\log _{a} y-2\right) \log _{18} a=1, \\ 2 x+y-20 a=0 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ y>0, \\ 0<a \neq 1 .\end{array}\right.$
From the first equation of the system, we get $\log _{a} \frac{x y}{a^{2}}=\log _{a} 18 \Leftrightarrow$ $\Leftrightarrow \frac{x y}{a^{2}}=18, x y=18 a^{2}$. The system has the form $\left\{\begin{array}{l}x y=... | (18),(92) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,449 |
7.275.
$$
\left\{\begin{array}{l}
(x+y) \cdot 3^{y-x}=\frac{5}{27} \\
3 \log _{5}(x+y)=x-y
\end{array}\right.
$$ | Solution.
Domain of definition: $x+y>0$.
Taking the logarithm of the first equation to the base 5, we have
$\log _{5}(x+y) \cdot 3^{y-x}=\log _{5} \frac{5}{27} \Leftrightarrow \log _{5}(x+y)-(x-y) \log _{5} 3=1-3 \log _{5} 3$.
We obtain the system
$$
\begin{aligned}
& \left\{\begin{array}{l}
\log _{5}(x+y)-(x-y) \... | (4;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,450 |
7.276. $\left\{\begin{array}{l}2^{\sqrt{x y}-2}+4^{\sqrt{x y}-1}=5, \\ \frac{3(x+y)}{x-y}+\frac{5(x-y)}{x+y}=8 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x y \geq 0, \\ x \neq \pm y .\end{array}\right.$
Transform the first equation of the system $2^{2 \sqrt{x y}}+2^{\sqrt{x y}}-20=0$ and, solving it as a quadratic equation in terms of $2^{\sqrt{x y}}$, we get $2^{\sqrt{2 x y}}=-5$ (no solutions), or $2^{\sqrt{... | (-4,-1),(4,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,451 |
7.277. $\left\{\begin{array}{l}x^{y}=2, \\ (2 x)^{y^{2}}=64(x>0)\end{array}\right.$ | ## Solution.
Domain of definition: $0<x \neq 1$.
Taking the logarithm of both equations of the system to the base 2, we get
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ \operatorname { log } _ { 2 } x ^ { y } = \operatorname { log } _ { 2 } 2 , } \\
{ \operatorname { log } _ { 2 } ( 2 x ) ^ { y ^ { 2 } } = \op... | (\frac{1}{\sqrt[3]{2}};-3),(\sqrt{2};2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,452 |
7.278.
$$
\left\{\begin{array}{l}
\frac{x^{2}}{y}+\frac{y^{2}}{x}=12 \\
2^{-\log _{2} x}+5^{\log _{s} \frac{1}{y}}=\frac{1}{3}
\end{array}\right.
$$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ y>0 .\end{array}\right.$
Transform the first equation of the system
$x^{3}+y^{3}=12 x y \Leftrightarrow(x+y)\left((x+y)^{2}-3 x y\right)=12 x y$.
The second equation, using the equality $a^{\log _{a} b}=b$, can be represented as $\frac{1}{x}+\frac{1}{y... | (6;6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,453 |
7.279. $\left\{\begin{array}{l}x^{y^{2}-7 y+10}=1, \\ x+y=8\end{array},(x>0)\right.$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
7.279. $\left\{\begin{array}{l}x^{y^{2}-7 y+10}=1, \\ x+y=8\end{array},(x>0)\right.$. | ## Solution.
Let's write the given system of equations as $\left\{\begin{array}{l}x^{y^{2}-7 y+10}=x^{0} \\ x+y=8 .\end{array}\right.$, This system of equations is equivalent to two systems:
1) $\left\{\begin{array}{l}x=1 \\ x+y=8\end{array} \Leftrightarrow\left\{\begin{array}{l}x=1 \\ y=7\end{array}\right.\right.$
2... | (1;7),(6;2),(3;5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,454 |
7.280. $\left\{\begin{array}{l}2\left(\log _{1 / y} x-2 \log _{x^{2}} y\right)+5=0, \\ x y^{2}=32 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}0<x \neq 1, \\ 0<y \neq 1 .\end{array}\right.$
In the first equation of the system, we switch to base 2, and in the second equation, we take the logarithm to base 2. We have
$\left\{\begin{array}{l}2\left(-\frac{\log _{2} x}{\log _{2} y}-\frac{\log _{2} y}{\... | (2;4),(4\sqrt{2};2\sqrt[4]{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,455 |
7.281. $\left\{\begin{array}{l}y \cdot x^{\log _{y} x}=x^{2.5} \\ \log _{3} y \cdot \log _{y}(y-2 x)=1 .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}0<x \neq 1, \\ 0<y \neq 1 .\end{array}\right.$
Logarithmize the first equation of the system with base 3, we have
$$
\begin{aligned}
& \log _{3}\left(y \cdot x^{\log _{y} x}\right)=\log _{3} x^{2.5} \Leftrightarrow \log _{3} y+\log _{3} x^{\log _{y} x}=2.5 \log... | (3;9) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,456 |
7.282. $\left\{\begin{array}{l}\lg (x-3)-\lg (5-y)=0, \\ 4^{-1} \cdot \sqrt[y]{4^{x}}-8 \sqrt[x]{8^{y}}=0 .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x>3 \\ 0 \neq y<5 .\end{array}\right.$
Rewrite the system of equations as
$\left\{\begin{array}{l}\lg \frac{x-3}{5-y}=0, \\ 2^{-2+\frac{2 x}{y}}=2^{3+\frac{3 y}{x}}\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{x-3}{5-y}=1, \\ \frac{2 x}{y}-2=3+\frac{3... | (6;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,457 |
7.283. $\left\{\begin{array}{l}\log _{x}(3 x+2 y)=2, \\ \log _{y}(2 x+3 y)=2 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}00, \\ 2 x+3 y>0 .\end{array}\right.$
Transform the system taking into account the domain of definition
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 3 x + 2 y = x ^ { 2 } , } \\
{ 2 x + 3 y = y ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}... | (5;5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,458 |
7.284. $\left\{\begin{array}{l}x+y=12, \\ 2\left(2 \log _{y^{2}} x-\log _{1 / x} y\right)=5 .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}0<x \neq 1, \\ 0<y \neq 1\end{array}\right.$
In the second equation of the system, we switch to base $y$. We have $2\left(\log _{y} x+\frac{1}{\log _{y} x}\right)=5 \Leftrightarrow 2 \log _{y}^{2} x-5 \log _{y} x+2=0 \Rightarrow \log _{y} x=\frac{1}{2}$ or $\log... | (3;9),(9;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,459 |
7.285. $\left\{\begin{array}{l}x^{x^{2}-y^{2}-16}=1, \\ x-y=2\end{array} \quad(x>0)\right.$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
7.285. $\left\{\begin{array}{l}x^{x^{2}-y^{2}-16}=1, \\ x-y=2\end{array} \quad(x>0)\right.... | ## Solution.
The first equation is equivalent to two equations: $x=1$ or $x^{2}-y^{2}-16=0$ for $x>0$. Then the system of equations is equivalent to two systems:
1) $\left\{\begin{array}{l}x=1, \\ x-y=2\end{array} \Leftrightarrow\left\{\begin{array}{l}x=1 \\ y=-1\end{array}\right.\right.$
2) $\left\{\begin{array}{l}x... | (1,-1),(5,3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,460 |
7.286. $\left\{\begin{array}{l}\lg \sqrt{(x+y)^{2}}=1, \\ \lg y-\lg |x|=\lg 2 .\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
\lg \sqrt{(x+y)^{2}}=1, \\
\lg y-\lg |x|=\lg 2 .
\end{array}\right.
\] | ## Solution.
Domain of definition: $\left\{\begin{array}{l}y>0 \\ x \neq 0 \\ x+y \neq 0 .\end{array}\right.$
Write the system of equations as $\left\{\begin{array}{l}\lg |x+y|=1, \\ \lg \frac{y}{|x|}=\lg 2\end{array} \Leftrightarrow\left\{\begin{array}{l}|x+y|=10, \\ \frac{y}{|x|}=2 .\end{array}\right.\right.$
This... | (-10,20),(\frac{10}{3},\frac{20}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,461 |
7.287. $\left\{\begin{array}{l}4^{x}-7 \cdot 2^{x-0.5 y}=2^{3-y}, \\ y-x=3 .\end{array}\right.$ | ## Solution.
From the second equation of the system, we find $y=x+3$. Then
$2^{2 x}-7 \cdot 2^{x-0.5(x+3)}=2^{3-x-3} \Leftrightarrow 2^{2 x}-7 \cdot 2^{0.5 x-1.5}=2^{-x} \Leftrightarrow$
$\Leftrightarrow 2^{3 x}-\frac{7 \cdot 2^{1.5 x}}{2^{1.5}}=1 \Leftrightarrow 2^{1.5} \cdot 2^{3 x}-7 \cdot 2^{1.5 x}-2^{1.5}=0$.
... | (1;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,462 |
7.288. $\left\{\begin{array}{l}5^{\sqrt[3]{x}} \cdot 2^{\sqrt{y}}=200 \\ 5^{2 \sqrt[3]{x}}+2^{2 \sqrt{y}}=689\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
5^{\sqrt[3]{x}} \cdot 2^{\sqrt{y}}=200 \\
5^{2 \sqrt[3]{x}}+2^{2 \sqrt{y}}=689
\end{array}\right.
\] | ## Solution.
Domain of definition: $y \geq 0$.
Rewrite the second equation of the system as
$$
\begin{aligned}
& \left(5^{\sqrt[3]{x}}+2^{\sqrt{y}}\right)^{2}-2 \cdot 5^{\sqrt[3]{x}} \cdot 2^{\sqrt{y}}=689 \Rightarrow\left(5^{\sqrt[3]{x}}+2^{\sqrt{y}}\right)^{2}=1089 \Rightarrow \\
& \Rightarrow 5^{\sqrt[3]{x}}+2^{\... | (27\log_{5}^{3}2;4\log_{2}^{2}5),(8;9) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,463 |
7.289. $\left\{\begin{array}{l}10^{\lg 0.5\left(x^{2}+y^{2}\right)+1.5}=100 \sqrt{10}, \\ \frac{\sqrt{x^{2}+10 y}}{3}=\frac{6}{2 \sqrt{x^{2}+10 y}-9}\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0, \\ 2 \sqrt{x^{2}+10 y}-9 \neq 0 .\end{array}\right.$
Rewrite the first equation of the system as
$$
\begin{aligned}
& 10^{\lg 0.5\left(x^{2}+y^{2}\right)+1.5}=10^{2.5} \Leftrightarrow \lg 0.5\left(x^{2}+y^{2}\right)+1.5=2.5 \Leftrighta... | (-4;2),(4;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,464 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.