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 |
|---|---|---|---|---|---|---|---|---|---|
6.091. $\left\{\begin{array}{l}(x-y) \cdot x y=30 \\ (x+y) \cdot x y=120\end{array}\right.$ | ## Solution.
After dividing the second equation of the system by the first, we get $\frac{(x+y) x y}{(x-y) x y}=\frac{120}{30}, y=\frac{3}{5} x$. From the first equation of the system, we find $\left(x-\frac{3}{5} x\right) x \cdot \frac{3}{5} x=30, x^{3}=5^{3}$, from which $x=5$; then $y=3$.
Answer: $(5 ; 3)$ | (5;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,748 |
6.093. $\left\{\begin{array}{l}v-u=1, \\ w-v=1, \\ (u-1)^{3}+(v-2)^{3}+(w-3)^{3}=3 .\end{array}\right.$ | ## Solution.
From the first equation of the system, we find $y=1+u$. Substituting this value of $v$ into the second and third equations of the system, we have
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ w - ( 1 + u ) = 1 , } \\
{ ( u - 1 ) ^ { 3 } + ( 1 + u - 2 ) ^ { 3 } + ( w - 3 ) ^ { 3 } = 3 }
\end{array} \... | (2;3;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,749 |
6.094. $\left\{\begin{array}{l}\frac{x+y}{x-y}+\frac{x-y}{x+y}=\frac{13}{6} \\ x y=5 .\end{array}\right.$ | ## Solution.
Domain of definition: $x \neq \pm y$.
Transforming the first equation of the system, we get
$$
6(x+y)^{2}+6(x-y)^{2}=13(x-y)(x+y) \Leftrightarrow x^{2}=25 y^{2}
$$
from which
$x_{1}=-5 y, x_{2}=5 y$.
From the second equation of the system, we find $y^{2}=-1$ (not suitable) or $y^{2}=1$, from which $y... | (5;1),(-5;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,750 |
6.095. $\left\{\begin{array}{l}3 x+2 y+2 z=13, \\ 2 x+3 y+2 z=14, \\ 2 x+2 y+3 z=15 .\end{array}\right.$ | ## Solution.
By adding all three equations, we get $7(x+y+z)=42$, from which $x+y+z=6$. Now we will sequentially subtract this equation from each equation in the system:
$$
\begin{aligned}
& \left\{\begin{array}{l}
3 x+2 y+2 z=13, \\
x+y+z=6
\end{array} \Leftrightarrow x=1 ;\left\{\begin{array}{l}
2 x+3 y+2 z=14 \\
x... | (1;2;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,751 |
6.096. $\left\{\begin{array}{l}x^{2} y^{3}=16 \\ x^{3} y^{2}=2\end{array}\right.$ | ## Solution.
Dividing the first equation of the system by the second, we get $\frac{y}{x}=8, y=8 x$.
From the second equation of the system, we have $x^{3} \cdot 64 x^{2}=2, x^{5}=\frac{1}{32}$, from which $x=\frac{1}{2};$ then $y=4$.
Answer: $\left(\frac{1}{2} ; 4\right)$ | (\frac{1}{2};4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,752 |
6.097. $\left\{\begin{array}{l}x+2 y+3 z=3, \\ 3 x+y+2 z=7, \\ 2 x+3 y+z=2 .\end{array}\right.$ | ## Solution.
We will transform the system using the Gaussian method, i.e., subtract the first equation, multiplied by the corresponding number, from the second and third equations.
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ x + 2 y + 3 z = 3 , } \\
{ 3 x + y + 2 z = 7 , } \\
{ 2 x + 3 y + z = 2 }
\end{array} ... | (2,-1,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,753 |
6.098. $\left\{\begin{array}{l}x^{3}+y^{3}=7 \\ x y(x+y)=-2\end{array}\right.$ | Solution.
Rewrite the system of equations as
$$
\left\{\begin{array} { l }
{ ( x + y ) ( x ^ { 2 } - x y + y ^ { 2 } ) = 7 , } \\
{ x y ( x + y ) = - 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x+y)\left((x+y)^{2}-3 x y\right)=7 \\
x y(x+y)=-2
\end{array}\right.\right.
$$
Let $\left\{\begin{array}{l}x+... | (2,-1),(-1,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,754 |
6.099. $\left\{\begin{array}{l}x^{2}+x y+y^{2}=91 \\ x+\sqrt{x y}+y=13\end{array}\right.$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6.099. $\left\{\begin{array}{l}x^{2}+x y+y^{2}=91 \\ x+\sqrt{x y}+y=13\end{array}\right... | Solution.
Domain of definition: $x y \geq 0$.
Let $\left\{\begin{array}{l}\sqrt{x}=u \geq 0, \\ \sqrt{y}=v \geq 0,\end{array}\right.$ then $\left\{\begin{array}{l}x=u^{2}, \\ y=v^{2},\end{array}\left\{\begin{array}{l}x^{2}=u^{4}, \\ y^{2}=v^{4} .\end{array}\right.\right.$
With respect to $u$ and $v$, the system of e... | (1;9)(9;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,755 |
6.102. $\left\{\begin{array}{l}\sqrt{2 x-y+11}-\sqrt{3 x+y-9}=3 \\ \sqrt[4]{2 x-y+11}+\sqrt[4]{3 x+y-9}=3\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
\sqrt{2 x - y + 11} - \sqrt{3 x + y - 9} = 3 \\
\sqrt[4]{2 x - y + 11} + \sqrt[4]{3 x + y - 9} = 3
\end{array}\right.
\] | Solution.
Let $\left\{\begin{array}{l}\sqrt[4]{2 x-y+11}=u \geq 0, \\ \sqrt[4]{3 x+y-9}=v \geq 0 .\end{array}\right.$
With respect to $u$ and $v$, the system takes the form
$$
\left\{\begin{array} { l }
{ u ^ { 2 } - v ^ { 2 } = 3 , } \\
{ u + v = 3 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ ( u - ... | (3;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,756 |
6.103. $\left\{\begin{array}{l}\sqrt{\frac{y}{x}}-2 \sqrt{\frac{x}{y}}=1, \\ \sqrt{5 x+y}+\sqrt{5 x-y}=4 .\end{array}\right.$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6.103. $\left\{\begin{array}{l}\sqrt{\frac{y}{x}}-2... | Solution.
Given: $\left\{\begin{array}{l}5 x+y \geq 0, \\ 5 x-y \geq 0, \\ \frac{y}{x}>0 .\end{array}\right.$
Let $\sqrt{\frac{y}{x}}=z>0$. Then $z-\frac{2}{z}=1$ or $z^{2}-z-2=0$, where $z \neq 0$. From this, $z_{1}=-1, z_{2}=2 ; z_{1}=-1<0$ is not suitable.
Then $\sqrt{\frac{y}{x}}=2, \frac{y}{x}=4, y=4 x$. From t... | (1;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,757 |
6.106. $\left\{\begin{array}{l}u^{2}+v^{2}=u v+13 \\ u+v=\sqrt{u v}+3\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
u^{2}+v^{2}=u v+13 \\
u+v=\sqrt{u v}+3
\end{array}\right.
\] | ## Solution.
Domain of definition: $u v \geq 0$.
Let $\left\{\begin{array}{l}\sqrt{u}=x \geq 0, \\ \sqrt{v}=y \geq 0,\end{array}\left\{\begin{array}{l}u=x^{2}, \\ v=y^{2},\end{array}\left\{\begin{array}{l}u^{2}=x^{4}, \\ v^{2}=y^{4} .\end{array}\right.\right.\right.$
With respect to $x$ and $y$, we obtain the system... | (1;4),(4;1),(2-\sqrt{3};2+\sqrt{3}),(2+\sqrt{3};2-\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,758 |
6.108. $\left\{\begin{array}{l}3(2-\sqrt{x-y})^{-1}+10(2+\sqrt{x+y})^{-1}=5 . \\ 4(2-\sqrt{x-y})^{-1}-5(2+\sqrt{x+y})^{-1}=3 .\end{array}\right.$
## Solution.
Rewrite the system of equations as
$\left\{\begin{array}{l}\frac{3}{2-\sqrt{x-y}}+\frac{10}{2+\sqrt{x+y}}=5 \\ \frac{4}{2-\sqrt{x-y}}-\frac{5}{2+\sqrt{x+y}}=3... | ## Solution.
Let $\left\{\begin{array}{l}\sqrt[3]{x}=u, \\ \sqrt[3]{y}=v,\end{array}\left\{\begin{array}{l}x=u^{3} \\ y=v^{3}\end{array}\right.\right.$, Relative to $u$ and $v$ the system takes the form
$\left\{\begin{array}{l}u+v=4, \\ u^{3}+v^{3}=28\end{array} \Leftrightarrow\left\{\begin{array}{l}u+v=4, \\ (u+v)\l... | (1;27),(27;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,759 |
6.111. $\left\{\begin{array}{l}2(\sqrt{x}+\sqrt{y})=3 \sqrt{x y} \\ x+y=5\end{array}\right.$
The system of equations is:
$\left\{\begin{array}{l}2(\sqrt{x}+\sqrt{y})=3 \sqrt{x y} \\ x+y=5\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \geq 0, \\ y \geq 0 .\end{array}\right.$
Rewrite the system of equations as $\left\{\begin{array}{l}2(\sqrt{x}+\sqrt{y})=3 \sqrt{x y}, \\ (\sqrt{x}+\sqrt{y})^{2}-\sqrt{x y}=5\end{array}\right.$ and introduce the substitution $\left\{\begin{array}{l}\sqrt{x}... | (4;1)(1;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,760 |
6.112. $\left\{\begin{array}{l}\sqrt{x}+\sqrt{y}=10 \\ \sqrt[4]{x}+\sqrt[4]{y}=4\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x \geq 0, \\ y \geq 0 .\end{array}\right.$
Let $\left\{\begin{array}{l}\sqrt[4]{x}=u \geq 0, \\ \sqrt[4]{y}=v \geq 0,\end{array}\left\{\begin{array}{l}\sqrt{x}=u^{2}, \\ \sqrt{y}=v^{2} .\end{array}\right.\right.$ Relative to $u$ and $v$, the system has the form
... | (1;81),(81;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,761 |
6.113. $\left\{\begin{array}{l}\sqrt{\frac{x+a}{y}}+\sqrt{\frac{y}{x+a}}=2, \\ x+y=x y+a .\end{array}\right.$ | ## Solution.
Domain of definition: $\frac{x+a}{y}>0$.
Let $\sqrt{\frac{x+a}{y}}=t$, where $t>0$. With respect to $t$, the equation takes the form $t+\frac{1}{t}=2, t^{2}-2 t+1=0, (t-1)^{2}=0$, from which $t=1$. Then $\sqrt{\frac{x+a}{y}}=1$, $\frac{x+a}{y}=1$, from which $y=x+a$. From the second equation, we get
$$
... | x_{1}=0,y_{1}=,x_{2}=2-,y_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,762 |
6.114. $\left\{\begin{array}{l}y \sqrt{2 x}-x \sqrt{2 y}=6, \\ x y^{2}-x^{2} y=30\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \geq 0, \\ y \geq 0 .\end{array}\right.$
Rewrite the first equation as $\sqrt{2}\left(\sqrt{x y^{2}}-\sqrt{x^{2} y}\right)=6$ and let $\sqrt{x y^{2}}=u, \sqrt{x^{2} y}=v$, where $u \geq 0, v \geq 0$. Then the system in terms of $u$ and $v$ becomes
$$
\begi... | (\frac{1}{2};8) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,763 |
6.115. $\left\{\begin{array}{l}\sqrt[3]{x}+\sqrt[3]{y}=3 \\ \sqrt[3]{x^{2}}-\sqrt[3]{x y}+\sqrt[3]{y^{2}}=3\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
\sqrt[3]{x} + \sqrt[3]{y} = 3 \\
\sqrt[3]{x^{2}} - \sqrt[3]{x y} + \sqrt[3]{y^{2}} = 3
\end{array}\right.
\] | Solution.
Let $\left\{\begin{array}{l}\sqrt[3]{x}=u, \\ \sqrt[3]{y}=v,\end{array}\left\{\begin{array}{l}x=u^{3}, \\ y=v^{3}\end{array}\right.\right.$
With respect to $u$ and $v$, the system takes the form
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ u + v = 3 , } \\
{ u ^ { 2 } - u v + v ^ { 2 } = 3 }
\end{arr... | (1;8),(8;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,764 |
6.116. $\left\{\begin{array}{l}\sqrt[4]{u}-\sqrt[4]{v}=1 \\ \sqrt{u}+\sqrt{v}=5\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}u \geq 0, \\ v \geq 0 .\end{array}\right.$
Let $\left\{\begin{array}{l}\sqrt[4]{u}=x \geq 0, \\ \sqrt[4]{v}=y \geq 0 .\end{array}\right.$ Relative to $x$ and $y$, the system takes the form
$\left\{\begin{array}{l}x-y=1, \\ x^{2}+y^{2}=5\end{array} \Leftrightarr... | (16;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,765 |
6.118. $\left\{\begin{array}{l}\sqrt{\frac{x+y}{2}}+\sqrt{\frac{x-y}{3}}=14, \\ \sqrt{\frac{x+y}{8}}-\sqrt{\frac{x-y}{12}}=3 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x+y \geq 0, \\ x-y \geq 0 .\end{array}\right.$
Let $\left\{\begin{array}{l}\sqrt{\frac{x+y}{2}}=u, \\ \sqrt{\frac{x-y}{3}}=v,\end{array}\right.$ where $u \geq 0$ and $v \geq 0$.
In terms of $u$ and $v$, the system becomes $\left\{\begin{array}{l}u+v=14, \\ \... | (124;76) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,766 |
6.119. $\left\{\begin{array}{l}\sqrt{x}-\sqrt{y}=0.5 \sqrt{x y} \\ x+y=5 .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x \geq 0 \\ y \geq 0 .\end{array}\right.$
Let $\left\{\begin{array}{l}\sqrt{x}=u, \\ \sqrt{y}=v,\end{array}\right.$ where $u \geq 0$ and $v \geq 0$.
With respect to $u$ and $v$, the system takes the form
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ u - v... | (4;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,767 |
6.122. Form a second-degree equation, one of whose roots is equal to the sum, and the other to the product of the roots of the equation $a x^{2}+b x+c=0$. | Solution.
Let $y^{2}+p y+q=0$ be the desired equation with roots $y_{1}=x_{1}+x_{2}$, $y_{2}=x_{1} \cdot x_{2}$. By Vieta's theorem, we have
$$
\left\{\begin{array} { l }
{ y _ { 1 } = x _ { 1 } + x _ { 2 } = - \frac { b } { a } } \\
{ y _ { 2 } = x _ { 1 } \cdot x _ { 2 } = \frac { c } { a } }
\end{array} \quad \te... | ^{2}y^{2}+(b-)y-=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,768 |
6.123. Form a quadratic equation whose roots are one unit greater than the roots of the equation $a x^{2}+b x+c=0$. | Solution.
Let $y^{2}+p y+q=0$ be the desired equation with roots $y_{1}=x_{1}+1$, $y_{2}=x_{2}+1$. From the condition by Vieta's theorem
$$
\begin{aligned}
& \left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{b}{a}, \\
x_{1} \cdot x_{2}=\frac{c}{a},
\end{array}\right. \\
& \left\{\begin{array}{l}
y_{1}+y_{2}=x_{1}+1+x_{2}+1=... | ^{2}+(b-2)y+-b+=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,769 |
6.124. Determine the coefficients of the quadratic equation
$x^{2}+p x+q=0$
so that its roots are equal to $p$ and $q$. | ## Solution.
By Vieta's theorem $\left\{\begin{array}{l}p+q=-p, \\ p q=q\end{array} \Leftrightarrow\left\{\begin{array}{l}2 p+q=0, \\ q(p-1)=0 .\end{array}\right.\right.$
From the second equation of the system, we have $q=0$ or $p-1=0$. Then $q_{1}=0, p_{1}=0 ; p_{2}=1, q_{2}=-2 p_{2}=-2$.
Answer: $p_{1}=q_{1}=0 ; p... | p_{1}=q_{1}=0;p_{2}=1,q_{2}=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,770 |
6.125. Find the coefficients $A$ and $B$ of the equation $x^{2}+A x+B=0$, if it is known that the numbers $A$ and $B$ are its roots. | ## Solution.
By Vieta's theorem
$$
\left\{\begin{array} { l }
{ A + B = - A , } \\
{ A B = B }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ 2 A + B = 0 , } \\
{ B ( A - 1 ) = 0 }
\end{array} \Rightarrow \left\{\begin{array} { l }
{ A _ { 1 } = 0 , } \\
{ B _ { 1 } = 0 }
\end{array} \text { or } \left\{... | A_{1}=B_{1}=0;A_{2}=1,B_{2}=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,771 |
6.126. For what integer value of $k$ is one of the roots of the equation $4 x^{2}-(3 k+2) x+\left(k^{2}-1\right)=0$ three times smaller than the other? | Solution.
From the condition, by Vieta's theorem, we have
where $k \in \mathbb{Z}$. From this, $37 k^{2}-36 k-76=0, k_{1}=2, k_{2}=-\frac{38}{37} \notin \mathbb{Z}$ (does not fit).
Answer... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,772 |
6.127. For what integer value of $p$ do the equations $3 x^{2}-4 x+p-2=0$ and $x^{2}-2 p x+5=0$ have a common root? Find this root. | ## Solution.
Let $x_{1}$ be the common root, then
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 3 x _ { 1 } ^ { 2 } - 4 x _ { 1 } + p - 2 = 0 , } \\
{ x _ { 1 } ^ { 2 } - 2 p x _ { 1 } + 5 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
3 x_{1}^{2}-4 x_{1}+p-2=0, \\
3 x_{1}^{2}-6 p x_{1}+15=0
\end{arra... | 1,p=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,773 |
6.128. Find all values of $a$ for which the sum of the roots of the equation $x^{2}-2 a(x-1)-1=0$ is equal to the sum of the squares of the roots. | ## Solution.
Given $x^{2}-2 a x+(2 a-1)=0$. By Vieta's theorem $\left\{\begin{array}{l}x_{1}+x_{2}=2 a, \\ x_{1} \cdot x_{2}=2 a-1 .\end{array}\right.$ Further,
$$
x_{1}+x_{2}=x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} \Leftrightarrow x_{1}+x_{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} .
$$
Us... | a_{1}=\frac{1}{2},a_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,774 |
6.129. For what value of $a$ do the equations $x^{2}+a x+8=0$ and $x^{2}+x+a=0$ have a common root? | ## Solution.
Let $x_{1}$ be the common root, then
$$
\left\{\begin{array}{l}
x_{1}^{2}+a x_{1}+8=0, \\
x_{1}^{2}+x_{1}+a=0
\end{array} \Rightarrow a x_{1}-x_{1}+8-a=0, x_{1}=\frac{a-8}{a-1}\right.
$$
From the second equation of the system, we have
$$
\begin{aligned}
& \left(\frac{a-8}{a-1}\right)^{2}+\left(\frac{a-... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,775 |
6.130. In the equation $x^{2}-2 x+c=0$, determine the value of $c$ for which its roots $x_{1}$ and $x_{2}$ satisfy the condition $7 x_{2}-4 x_{1}=47$. | Solution.
From the condition by Vieta's theorem we have $\left\{\begin{array}{l}x_{1}+x_{2}=2, \\ x_{1} \cdot x_{2}=c, \\ 7 x_{2}-4 x_{1}=47 .\end{array}\right.$ From here, $x_{2}=2-x_{1}$ and we obtain
$$
\left\{\begin{array} { l }
{ x _ { 1 } ( 2 - x _ { 1 } ) = c } \\
{ 7 ( 2 - x _ { 1 } ) - 4 x _ { 1 } = 4 7 }
\... | -15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,776 |
6.131. Without solving the equation $x^{2}-(2 a+1) x+a^{2}+2=0$, find the value of $a$ for which one of the roots is twice the other. | Solution.
From the condition, by Vieta's theorem, we have
$$
\left\{\begin{array} { l }
{ x _ { 1 } + x _ { 2 } = 2 a + 1 , } \\
{ x _ { 1 } \cdot x _ { 2 } = a ^ { 2 } + 2 , } \\
{ x _ { 2 } = 2 x _ { 1 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ 3 x _ { 1 } = 2 a + 1 , } \\
{ 2 x _ { 1 } ^ { 2 } ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,777 |
6.132. For what value of $p$ is the ratio of the roots of the equation $x^{2}+p x-16=0$ equal to $-4$? | ## Solution.
By Vieta's theorem and the condition, we have the system
$\left\{\begin{array}{l}x_{1}+x_{2}=-p, \\ x_{1} \cdot x_{2}=-16, \\ \frac{x_{2}}{x_{1}}=-4\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{1}+x_{2}=-p, \\ x_{1} \cdot x_{2}=-16, \\ x_{2}=-4 x_{1}\end{array} \Rightarrow\left\{\begin{array}{l}x_... | p_{1,2}=\6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,778 |
6.133. Without solving the equation $3 x^{2}-5 x-2=0$, find the sum of the cubes of its roots. | ## Solution.
By Vieta's theorem and the given condition, we have the system
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=\frac{5}{3} \\
x_{1} \cdot x_{2}=-\frac{2}{3}
\end{array}\right.
$$
and
$x_{1}^{3}+x_{2}^{3}=\left(x_{1}+x_{2}\right)\left(x_{1}^{2}-x_{1} x_{2}+x_{2}^{3}\right)=\left(x_{1}+x_{2}\right)\left(\left(x_{... | \frac{215}{27} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,779 |
6.134. For what integer value of $b$ do the equations $2 x^{2}+(3 b-1) x-3=0$ and $6 x^{2}-(2 b-3) x-1=0$ have a common root? | Solution.
Let $x_{1}$ be the common root. Then
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 2 x _ { 1 } ^ { 2 } + ( 3 b - 1 ) x _ { 1 } - 3 = 0 , } \\
{ 6 x _ { 1 } ^ { 2 } - ( 2 b - 3 ) x _ { 1 } - 1 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
6 x^{2}+(9 b-3) x-9=0, \\
6 x^{2}-(2 b-3) x_{1}-1=0
\... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,780 |
7.001. $\sqrt{25^{\frac{1}{\log _{6} 5}}+49^{\frac{1}{\log _{8} 7}}}$ | ## Solution.
$$
\begin{aligned}
& \sqrt{\frac{1}{25^{\log _{6} 5}}+49^{\frac{1}{\log _{8} 7}}}=\sqrt{5^{2 \log _{5} 6}+7^{2 \log _{7} 8}}=\sqrt{5^{\log _{5} 6^{2}}+7^{\log _{7} 8^{2}}}= \\
& =\sqrt{6^{2}+8^{2}}=10
\end{aligned}
$$
Answer: 10. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,782 |
7.002. $81^{\frac{1}{\log _{5} 3}}+27^{\log _{9} 36}+3^{\frac{4}{\log _{7} 9}}$. | Solution.
$81^{\frac{1}{\log _{5} 3}}+27^{\log _{9} 36}+3^{\frac{4}{\log _{7} 9}}=3^{4 \log _{3} 5}+3^{\frac{3}{2^{2} \log _{3} 36}}+3^{\frac{4}{2} \log _{3} 7}=5^{4}+36^{\frac{3}{2}}+49=$ $=625+216+49=890$.
Answer: 890. | 890 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,783 |
7.003. $-\log _{2} \log _{2} \sqrt{\sqrt[4]{2}}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
7.003. $-\log _{2} \log _{2} \sqrt{\sqrt[4]{2}}$. | Solution.
$$
-\log _{2} \log _{2} \sqrt{\sqrt[4]{2}}=-\log _{2} \log _{2} 2^{\frac{1}{8}}=-\log _{2} \frac{1}{8} \log _{2} 2=-\log _{2} 2^{-3}=3
$$
Answer: 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,784 |
7.004. $-\log _{3} \log _{3} \sqrt[3]{\sqrt[3]{3}}$.
7.004. $-\log _{3} \log _{3} \sqrt[3]{\sqrt[3]{3}}$. | ## Solution.
$-\log _{3} \log _{3} \sqrt[3]{\sqrt[3]{3}}=-\log _{3} \log _{3} 3^{\frac{1}{9}}=-\log _{3} \frac{1}{9} \log _{3} 3=-\log _{3} 3^{-2}=2$.
Answer: 2.
## Solution.
 \cdot 49^{\log _{7} 2}$. | Solution.
$$
\begin{aligned}
& \left(81^{\frac{1}{4}-\frac{1}{2} \log _{9} 4}+25^{\log _{125} 8}\right) \cdot 49^{\log _{7} 2}=\left(\frac{81^{\frac{1}{4}}}{\left(9^{2}\right)^{\frac{1}{2} \log _{9} 4}}+5^{2 \log _{5} 32^{3}}\right) \cdot 7^{2 \log _{7} 2}= \\
& =\left(\frac{3}{4}+4\right) \cdot 4=19
\end{aligned}
$$
... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,786 |
7.008. $\frac{81^{\frac{1}{\log _{5} 9}}+3^{\frac{3}{\log _{\sqrt{6}} 3}}}{409} \cdot\left((\sqrt{7})^{\frac{2}{\log _{25} 7}}-125^{\log _{25} 6}\right)$
| ## Решение.
$$
\frac{81^{\frac{1}{\log _{5} 9}}+3^{\frac{3}{\log _{\sqrt{6}}}}}{409} \cdot\left((\sqrt{7})^{\frac{2}{\log _{25} 7}}-125^{\log _{25} 6}\right)=
$$
$=\frac{9^{2 \log _{9} 5}+3^{3 \log _{3} \sqrt{6}}}{409} \cdot\left(\left(7^{\frac{1}{2}}\right)^{2 \log _{7} 25}-5^{3 \log _{5} 26}\right)=\frac{9^{\log _{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,787 |
7.009. $\left(N^{\frac{1}{\log _{2} N}} \cdot N^{\frac{1}{\log _{4} N}} \cdot N^{\frac{1}{\log _{8} N}} \cdots N^{\frac{1}{\log _{64} N}}\right)^{\frac{1}{15}}$ (the bases of the logarithms are consecutive natural powers of the number 2). | ## Solution.
$$
\begin{aligned}
& \left(N^{\frac{1}{\log _{2} N}} \cdot N^{\frac{1}{\log _{4} N}} \cdot N^{\frac{1}{\log _{8} N}} \cdots N^{\frac{1}{\log _{512} N}}\right)^{\frac{1}{15}}= \\
& =\left(N^{\log _{N} 2} \cdot N^{\log _{N} 4} \cdot N^{\log _{N} 8} \cdots N^{\log _{N} 512}\right)^{\frac{1}{15}}= \\
& =(2 \c... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,788 |
7.010. $\left(2^{\log _{\sqrt{2}} a}-3^{\log _{27}\left(a^{2}+1\right)^{\beta}}-2 a\right):\left(7^{4 \log _{49} a}-5^{0.5 \log _{\sqrt{5}} a}-1\right)$ | ## Solution.
$$
\begin{aligned}
& \left(2^{\log _{\sqrt{2}} a}-3^{\log _{27}\left(a^{2}+1\right)^{1}}-2 a\right):\left(7^{4 \log _{49} a}-5^{0.5 \log _{\sqrt{5}} a}-1\right)= \\
& =\left(2^{\log _{2} a^{4}}-3^{\log _{3}\left(a^{2}+1\right)}-2 a\right):\left(7^{\log _{7} a^{2}}-5^{\log _{5} a}-1\right)= \\
& =\left(a^{... | ^{2}++1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,789 |
7.011. $\frac{\log _{a} \sqrt{a^{2}-1} \cdot \log _{1 / a}^{2} \sqrt{a^{2}-1}}{\log _{a^{2}}\left(a^{2}-1\right) \cdot \log _{\sqrt[3]{a}} \sqrt[6]{a^{2}-1}}$. | Solution.
$$
\frac{\log _{a} \sqrt{a^{2}-1} \cdot \log _{1 / a}^{2} \sqrt{a^{2}-1}}{\log _{a^{2}}\left(a^{2}-1\right) \cdot \log _{\sqrt[3]{a}} \sqrt[6]{a^{2}-1}}=\frac{\frac{1}{2} \log _{a}\left(a^{2}-1\right) \cdot \frac{1}{4} \log _{a}^{2}\left(a^{2}-1\right)}{\frac{1}{2} \log _{a}\left(a^{2}-1\right) \cdot \frac{1... | \log_{}\sqrt{^{2}-1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,790 |
7.013. $\frac{\left(25^{\frac{1}{2 \log _{4} 25}}+2 \log _{2} \log _{2} \log _{2} a^{2 \log _{a} 4}\right) \cdot 4^{-\frac{2}{\log _{3} 4}}-a^{2}}{1-a}$. | ## Solution.
$$
\frac{\left(25^{\frac{1}{2 \log _{9} 25}}+2 \log _{2} \log _{2} \log _{2} a^{2 \log _{a} 4}\right) \cdot 4^{\frac{2}{\log _{3} 4}}-a^{2}}{1-a}=
$$
$$
\begin{aligned}
& =\frac{\left(\left(25^{\log _{2} 49}\right)^{\frac{1}{2}}+2 \log _{2} \log _{2} 4\right) \cdot\left(\left(4^{\log _{4} 3}\right)^{-1}-... | 1+ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,792 |
7.014. $\left(\log _{a} b+\log _{b} a+2\right)\left(\log _{a} b-\log _{a b} b\right) \log _{b} a-1$. | Solution.
$$
\begin{aligned}
& \left(\log _{a} b+\log _{b} a+2\right)\left(\log _{a} b-\log _{a b} b\right) \log _{b} a-1=\left(\log _{a} b+\frac{1}{\log _{a} b}+2\right) \times \\
& \times\left(\log _{a} b-\frac{\log _{a} b}{\log _{a} a b}\right) \frac{1}{\log _{a} b}-1=\frac{\log _{a}^{2} b+2 \log _{a} b+1}{\log _{a... | \log_{}b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,793 |
7.015. $\frac{1-\log _{a}^{3} b}{\left(\log _{a} b+\log _{b} a+1\right) \cdot \log _{a} \frac{a}{b}}$. | ## Solution.
$\frac{1-\log _{a}^{3} b}{\left(\log _{a} b+\log _{b} a+1\right) \cdot \log _{a} \frac{a}{b}}=\frac{\left(1-\log _{a} b\right)\left(1+\log _{a} b+\log _{a}^{2} b\right)}{\left(\log _{a} b+\frac{1}{\log _{a} b}+1\right)\left(\log _{a} a-\log _{a} b\right)}=$
$$
=\frac{\left(1-\log _{a} b\right)\left(1+\lo... | \log_{}b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,794 |
7.016. If $\log _{a} 27=b$, then what is $\log _{\sqrt{3}} \sqrt[6]{a}$? | Solution.
$\log _{\sqrt{3}} \sqrt[6]{a}=\frac{1}{6} \cdot 2 \log _{3} a=\frac{1}{3 \log _{a} 3}=\frac{1}{\log _{a} 27}=\frac{1}{b}$.
Answer: $\frac{1}{b}$. | \frac{1}{b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,795 |
7.017. Show that under the conditions $x>0$ and $y>0$, from the equation $x^{2}+4 y^{2}=12 x y$ it follows that $\lg (x+2 y)-2 \lg 2=0.5(\lg x+\lg y)$. | ## Solution.
From the condition, we have: $(x+2 y)^{2}-2 x \cdot 2 y=12 x y, \quad(x+2 y)^{2}=16 x y$.
Taking the logarithm of both sides of the obtained equation with base 10, we get:
$$
\begin{aligned}
& \lg (x+2 y)^{2}=\lg 16 x y, \quad 2 \lg (x+2 y)=\lg 16+\lg x+\lg y \\
& 2 \lg (x+2 y)=4 \lg 2+\lg x+\lg y, \qua... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,796 |
7.019. Prove that if $y=2^{x^{2}}$ and $z=2^{y^{2}}$, then $x= \pm \sqrt{0.5 \log _{2} \log _{2} z}$, and specify all $z$ for which $x$ takes real values. | Solution.
Given $y>0$ and $z>0$. Taking the logarithm of both sides of the equation with base 2, we get $\log _{2} y=\log _{2} 2^{x^{2}}, \log _{2} y=x^{2}$, from which $x= \pm \sqrt{\log _{2} y}$. Similarly, $z=2^{y^{2}} \Rightarrow y=\sqrt{\log _{2} z}$.
Thus, $x= \pm \sqrt{\log _{2} \sqrt{\log _{2} z}}= \pm \sqrt{... | z\geq2 | Algebra | proof | Yes | Yes | olympiads | false | 50,797 |
7.020. $\left(1+\frac{1}{2 x}\right) \log 3+\log 2=\log \left(27-3^{1 / x}\right)$. | Solution.
D: $\quad\left\{\begin{array}{l}x \neq 0, \\ 27-3^{1 / x}>0 .\end{array}\right.$
$\lg 3^{1+\frac{1}{2 x}}+\lg 2=\lg \left(27-3^{\frac{1}{x}}\right) \lg \left(2 \cdot 3^{1+\frac{1}{2 x}}\right)=\lg \left(27-3^{\frac{1}{x}}\right)$
$2 \cdot 3^{1+\frac{1}{2 x}}=27-3^{\frac{1}{x}}, 3^{\frac{1}{x}}+6 \cdot 3^{\f... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,798 |
7.021. $3 \log _{5} 2+2-x=\log _{5}\left(3^{x}-5^{2-x}\right)$.
7.021. $3 \log _{5} 2+2-x=\log _{5}\left(3^{x}-5^{2-x}\right)$. | Solution.
Domain of definition: $3^{x}-5^{2-x}>0$.
$\log _{5} 8+2 \log _{5} 5-\log _{5}\left(3^{x}-25 \cdot 5^{-x}\right)=x \Leftrightarrow \log _{5} \frac{8 \cdot 25}{3^{x}-25 \cdot 5^{-x}}=x$,
from which $\frac{200}{3^{x}-25 \cdot 5^{-x}}=5^{x} \Leftrightarrow 15^{x}=15^{2}$. Therefore, $x=2$.
Answer: 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,799 |
7.022. $\sqrt{\log _{3} x^{9}}-4 \log _{9} \sqrt{3 x}=1$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ \log _{3} x>0, x>1\end{array}\right.$
$$
\begin{aligned}
& \sqrt{\log _{3} x^{9}}=1+4 \log _{9} \sqrt{3 x} \Leftrightarrow \sqrt{9 \log _{3} x}=1+\log _{3} 3 x \Leftrightarrow \\
& \Leftrightarrow \sqrt{9 \log _{3} x}=1+\log _{3} 3+\log _{3} x \Leftright... | 3;81 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,800 |
7.024. $\lg 5+\lg (x+10)=1-\lg (2 x-1)+\lg (21 x-20)$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x+10>0, \\ 2 x-1>0, \\ 21 x-20>0,\end{array} \quad x>\frac{20}{21}\right.$.
$$
\begin{aligned}
& \lg 5+\lg (x+10)=\lg 10-\lg (2 x-1)+\lg (21 x-20) \Leftrightarrow \lg 5(x+10)= \\
& =\lg \frac{10 \cdot(21 x-20)}{2 x-1} \Rightarrow 5(x+10)=\frac{10 \cdot(21 x-2... | 1.5;10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,801 |
7.028. $5^{2\left(\log _{5} 2+x\right)}-2=5^{x+\log _{5} 2}$. | ## Solution.
$\left(5^{x+\log _{5} 2}\right)^{2}-5^{x+\log _{5} 2}-2=0$; solving this equation as a quadratic equation in terms of $5^{x+\log _{5} 2}$, we find $5^{x+\log _{5} 2}=-1$ and $5^{x+\log _{5} 2}=2 ; 5^{x+\log _{5} 2}=-1$ has no solutions.
Thus,
$$
5^{x+\log _{5} 2}=2 \Rightarrow \log _{5} 5^{x+\log _{5} 2... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,802 |
7.029. $0.25^{\log _{2} \sqrt{x+3}-0.5 \log _{2}\left(x^{2}-9\right)}=\sqrt{2(7-x)}$. | ## Solution.
Domain of definition: $\quad\left\{\begin{array}{l}x+3>0, \\ x^{2}-9>0.33 . \Rightarrow x_{1}=5, x_{2}=-1 ; x_{2}=-1 \text{ does not fit the domain of definition.}\end{array}\right.$
Answer: 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,803 |
7.030. $x \lg \sqrt[5]{5^{2 x-8}}-\lg 25=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
7.030. $x \lg \sqrt[5]{5^{2 x-8}}-\lg 25=0$. | ## Solution.
$x \lg 5^{\frac{2 x-8}{5}}=\lg 25, \lg 5^{\frac{(2 x-8) x}{5}}=\lg 5^{2}, 5^{\frac{2 x^{2}-8 x}{5}}=5^{2}$,
$\frac{2 x^{2}-8 x}{5}=2, x^{2}-4 x-5=0$,
from which $x_{1}=5, x_{2}=-1$.
Answer: $5; -1$. | 5;-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,804 |
7.031. $\log _{5}(x-2)+\log _{\sqrt{5}}\left(x^{3}-2\right)+\log _{0.2}(x-2)=4$. | ## Solution.
Domain of definition: $\quad x-2>0, x>2$.
From the condition we have
$$
\log _{5}(x-2)+2 \log _{5}\left(x^{3}-2\right)-\log _{5}(x-2)=4, \log _{5}\left(x^{3}-2\right)=2
$$
from which $x^{3}-2=25, x^{3}=27$. Then $x=3$.
Answer: 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,805 |
7.034. $\lg \left(3^{x}-2^{4-x}\right)=2+0.25 \lg 16-0.5 x \lg 4$. | Solution.
Domain of definition: $3^{x}-2^{4-x}>0$.
From the condition
$$
\begin{aligned}
& \lg \left(3^{x}-2^{4-x}\right)=\lg 100+\lg 2-\lg 2^{x} \Rightarrow \lg \left(3^{x}-2^{4-x}\right)=\lg \frac{100 \cdot 2}{2^{x}} \\
& 3^{x}-2^{4-x}=\frac{200}{2^{x}}
\end{aligned}
$$
From here $6^{x}=216$, hence $x=3$.
Answer... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,806 |
7.035. $\log _{3}\left(81^{x}+3^{2 x}\right)=3 \log _{27} 90$. | ## Solution.
From the condition $\log _{3}\left(81^{x}+3^{2 x}\right)=\log _{3} 90, 9^{2 x}+9^{x}-90=0$, from which we find $9^{x}=-10$, which is not suitable, or $9^{x}=9$, from which we have $x=1$.
Answer: 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,807 |
7.036. $3 x-\log _{6} 8^{x}=\log _{6}\left(3^{3 x}+x^{2}-9\right)$
7.036. $3 x-\log _{6} 8^{x}=\log _{6}\left(3^{3 x}+x^{2}-9\right)$ | ## Solution.
Domain of definition: $3^{3 x}+x^{2}-9>0$.
From the condition $3 x=\log _{6} 8^{x}+\log _{6}\left(3^{3 x}+x^{2}-9\right) 3 x=\log _{6} 8^{x}\left(3^{3 x}+x^{2}-9\right)$, hence $6^{3 x}=8^{x}\left(3^{3 x}+x^{2}-9\right) 3^{3 x}=3^{3 x}+x^{2}-9 \Leftrightarrow x^{2}=9$. Then $x_{1,2}= \pm 3$.
Answer: $-3... | -3;3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,808 |
7.037. $\log _{6}\left(3^{x^{2}}+1\right)-\log _{6}\left(3^{2-x^{2}}+9\right)=\log _{6} 2-1$. | ## Solution.
From the condition
$$
\begin{aligned}
& \log _{6}\left(3^{x^{2}}+1\right)-\log _{6}\left(3^{2-x^{2}}+9\right)=\log _{6} 2-\log _{6} 6, \log _{6} \frac{3^{x^{2}}+1}{3^{2-x^{2}}+9}=\log _{6} \frac{2}{6} \\
& \frac{3^{x^{2}}+1}{9 \cdot 3^{-x^{2}}+9}=\frac{2}{6}, 3^{2 x^{2}}-2 \cdot 3^{x^{2}}-3=0 .
\end{alig... | -1;1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,809 |
7.038. $\lg \left(625 \sqrt[5]{5^{x^{2}-20 x+55}}\right)=0$.
7.038. $\lg \left(625 \sqrt[5]{5^{x^{2}-20 x+55}}\right)=0$. | ## Solution.
From the condition, we have $625 \cdot 5^{\frac{x^{2}-20 x+55}{5}}=1.5^{\frac{x^{2}-20 x+55}{5}}=5^{-4}$, from which $\frac{x^{2}-20 x+55}{5}=-4, x^{2}-20 x+75=0$. Then $x_{1}=5 ; x_{2}=15$.
Answer: $5 ; 15$. | 5;15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,810 |
7.039. $\lg \left(10^{\lg \left(x^{2}-21\right)}\right)-2=\lg x-\lg 25$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x^{2}-21>0, \\ x>0,\end{array} x>\sqrt{21}\right.$.
From the condition we have
$$
\lg \left(x^{2}-21\right)-\lg 100=\lg x-\lg 25, \lg \frac{x^{2}-21}{100}=\lg \frac{x}{25}, \quad \frac{x^{2}-21}{100}=\frac{x}{25}
$$
We obtain the quadratic equation $x^{2}-4... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,811 |
7.040. $\lg \left(x^{2}+1\right)=2 \lg ^{-1}\left(x^{2}+1\right)-1$.
7.040. $\lg \left(x^{2}+1\right)=2 \lg ^{-1}\left(x^{2}+1\right)-1$. | ## Solution.
Domain of definition: $x \neq 0$.
$$
\lg \left(x^{2}+1\right)=\frac{2}{\lg \left(x^{2}+1\right)}-1, \lg ^{2}\left(x^{2}+1\right)+\lg \left(x^{2}+1\right)-2=0
$$
Solving this equation as a quadratic equation in terms of $\lg \left(x^{2}+1\right)$, we find $\lg \left(x^{2}+1\right)=-2$ and $\lg \left(x^{2... | -3;3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,812 |
7.041. $\lg \sqrt{5^{x(13-x)}}+11 \lg 2=11$. | Solution.
$\lg 5^{\frac{x(13-x)}{2}}+\lg 2^{11}=11, \lg \left(5^{\frac{x(13-x)}{2}} \cdot 2^{11}\right)=11$
From here we have $5^{\frac{x(13-x)}{2}} \cdot 2^{11}=10^{11}, 5^{\frac{x(13-x)}{2}}=5^{11}$. Then $\frac{x(13-x)}{2}=11$, $x^{2}-13 x+22=0$, from which $x_{1}=2 ; x_{2}=11$.
Answer: $2 ; 11$. | 2;11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,813 |
7.042. $x(\lg 5-1)=\lg \left(2^{x}+1\right)-\lg 6$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
7.042. $x(\lg 5-1)=\lg \left(2^{x}+1\right)-\lg 6$. | Solution.
$$
\begin{aligned}
& x(\lg 5-\lg 10)=\lg \left(2^{x}+1\right)-\lg 6, \quad x \lg \frac{5}{10}=\lg \frac{2^{x}+1}{6} \\
& \lg 2^{-x}=\lg \frac{2^{x}+1}{6}, 2^{-x}=\frac{2^{x}+1}{6}, 2^{2 x}+2^{x}-6=0
\end{aligned}
$$
Solving this equation as a quadratic in terms of $2^{x}$, we find $2^{x}=-3$ (not valid), $2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,814 |
7.043. $\lg \left(81 \cdot \sqrt[3]{3^{x^{2}-8 x}}\right)=0$.
7.043. $\lg \left(81 \cdot \sqrt[3]{3^{x^{2}-8 x}}\right)=0$. | ## Solution.
We have $81 \cdot \sqrt[3]{3^{x^{2}-8 x}}=1, 3^{\frac{x^{2}-8 x}{3}}=3^{-4}$, from which $\frac{x^{2}-8 x}{3}=-4, x^{2}-8 x+12=0$; $x_{1}=2 ; x_{2}=6$.
Answer: $2 ; 6$. | 2;6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,815 |
7.044. $\log _{x} 9 x^{2} \cdot \log _{3}^{2} x=4$. | ## Solution.
Domain of definition: $\quad 0<x \neq 1$.
## We have
$$
\frac{\log _{3} 9 x^{2}}{\log _{3} x} \cdot \log _{3}^{2} x=4, \quad\left(\log _{3} 9+\log _{3} x^{2}\right) \log _{3} x=4, \log _{3}^{2} x+\log _{3} x-2=0
$$
Solving this equation as a quadratic equation in terms of $\log _{3} x$, we find $\left(... | \frac{1}{9};3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,816 |
7.045. $\log _{5}(3 x-11)+\log _{5}(x-27)=3+\log _{5} 8$. | Solution.
Domain of definition: $\left\{\begin{array}{l}3 x-11>0, \\ x-27>0,\end{array} \quad x>27\right.$.
## We have
$$
\begin{aligned}
& \log _{5}(3 x-11)+\log _{5}(x-27)=\log _{5} 125+\log _{5} 8 \\
& \log _{5}(3 x-11) \cdot(x-27)=\log _{5}(125 \cdot 8) \quad(3 x-11)(x-27)=125 \cdot 8 \\
& 3 x^{2}-92 x-703=0
\en... | 37 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,817 |
7.046. $\lg (5-x)+2 \lg \sqrt{3-x}=1$.
7.046. $\lg (5-x)+2 \lg \sqrt{3-x}=1$. | ## Solution.
Domain of definition: $\quad\left\{\begin{array}{l}5-x>0, \\ 3-x>0,\end{array}, x<3\right.$.
We have $\lg (5-x)+\lg (3-x)=1, \lg (5-x)(3-x)=1$, hence $(5-x)(3-x)=10$, $x^{2}-8 x+5=0$. Then $x_{1}=4-\sqrt{11}, x_{2}=4+\sqrt{11} ; x_{2}=4+\sqrt{11}$ does not satisfy the domain of definition.
Answer: $4-\s... | 4-\sqrt{11} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,818 |
7.047. Find the natural number $n$ from the equation
$$
3^{2} \cdot 3^{5} \cdot 3^{8} \cdots 3^{3 n-1}=27^{5}
$$ | ## Solution.
$3^{2+5+8+\ldots+3 n-1}=3^{15}, 2+5+8+\ldots+3 n-1=15$.
On the left side of the equation, we have the sum of the terms of an arithmetic progression $S_{k}$, where $a_{1}=2, d=3, a_{k}=3 n-1, k=\frac{a_{k}-a_{1}}{d}+1=\frac{3 n-1-2}{3}+1=n$.
Then $S_{k}=\frac{a_{1}+a_{k}}{2} \cdot k=\frac{2+3 n-1}{2} \cd... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,819 |
7.048. $0.5\left(\lg \left(x^{2}-55 x+90\right)-\lg (x-36)\right)=\lg \sqrt{2}$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x^{2}-55 x+90>0 \\ x-36>0\end{array}\right.$
From the condition
$$
\begin{aligned}
& 0.5\left(\lg \left(x^{2}-55 x+90\right)-\lg (x-36)\right)=0.5 \lg 2, \lg \frac{x^{2}-55 x+90}{x-36}=\lg 2 \\
& \frac{x^{2}-55 x+90}{x-36}=2
\end{aligned}
$$
We have $x^{2}-... | 54 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,820 |
7.049. $\lg (5-x)-\frac{1}{3} \lg \left(35-x^{3}\right)=0$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}5-x>0, \\ 35-x^{3}>0,\end{array}, x<\sqrt[3]{35}\right.$.
From the condition, we have $3 \lg (5-x)=\lg \left(35-x^{3}\right), \lg (5-x)^{3}=\lg \left(35-x^{3}\right)$, hence $(5-x)^{3}=35-x^{3}, x^{2}-5 x+6=0$. Therefore, $x_{1}=2, x_{2}=3$.
Answer: $2 ; 3$. | 2;3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,821 |
7.050. $\log _{2} \frac{x-5}{x+5}+\log _{2}\left(x^{2}-25\right)=0$. | ## Solution.
Domain of definition: $\frac{x-5}{x+5}>0$ or $x \in(-\infty ;-5) \cup(5 ; \infty)$.
We have $\log _{2} \frac{(x-5)\left(x^{2}-25\right)}{x+5}=0,(x-5)^{2}=1$, from which $x-5=-1$ or $x-5=1$. Then $x_{1}=4, x_{2}=6 ; x_{1}=4$ does not fit the domain of definition.
Answer: 6. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,822 |
7.051. $\frac{\lg 8-\lg (x-5)}{\lg \sqrt{x+7}-\lg 2}=-1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x-5>0, \\ x+7>0, \quad x>5 . \\ \sqrt{x+7} \neq 2,\end{array}\right.$
## From the condition
$$
\begin{aligned}
& \lg 8-\lg (x-5)=\lg 2-\lg \sqrt{x+7}, \quad \lg \frac{8}{x-5}=\lg \frac{2}{\sqrt{x+7}} \\
& \frac{8}{x-5}=\frac{2}{\sqrt{x+7}}, \quad 4 \sqrt{x+7... | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,823 |
7.052. $\log _{0.5}^{2} 4 x+\log _{2} \frac{x^{2}}{8}=8$. | ## Solution.
Domain of definition: $x>0$.
We have
$$
\begin{aligned}
& \log _{2}^{2} 4 x+\log _{2} \frac{x^{2}}{8}-8=0, \quad\left(\log _{2} 4+\log _{2} x\right)^{2}+\log _{2} x^{2}-\log _{2} 8-8=0 \\
& \log _{2}^{2} x+6 \log _{2} x-7=0
\end{aligned}
$$
Solving this equation as a quadratic equation in terms of $\lo... | \frac{1}{128};2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,824 |
7.053. $\lg (\lg x)+\lg \left(\lg x^{3}-2\right)=0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
7.053. $\lg (\lg x)+\lg \left(\lg x^{3}-2\right)=0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\lg x>0, \\ \lg x^{3}-2>0,\end{array} \quad x>\sqrt[3]{100}\right.$.
From the condition we have
$\lg \left(\lg x \cdot\left(\lg x^{3}-2\right)\right)=0, \quad \lg x(3 \lg x-2)=1$,
$3 \lg ^{2} x-2 \lg x-1=0$.
Solving this equation as a quadratic equation in te... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,825 |
7.054. $\log _{2} x+\log _{4} x+\log _{8} x=11$. | ## Solution.
Domain: $x>0$.
We have $\log _{2} x+\frac{1}{2} \log _{2} x+\frac{1}{3} \log _{2} x=11, \log _{2} x=6$, from which $x=2^{6}=64$.
Answer: 64. | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,826 |
7.055. $\log _{3}\left(3^{x}-8\right)=2-x$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
7.055. $\log _{3}\left(3^{x}-8\right)=2-x$. | Solution.
Domain of definition: $3^{x}-8>0$.
By the definition of logarithm, we have $3^{x}-8=3^{2-x}, 3^{x}-8=\frac{9}{3^{x}}$, $3^{2 x}-8 \cdot 3^{x}-9=0$, from which, solving this equation as a quadratic equation in terms of $3^{x}$, we find $3^{x}=-1, \varnothing$; or $3^{x}=9$, from which $x=2$.
Answer: 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,827 |
7.056. $7^{\lg x}-5^{\lg x+1}=3 \cdot 5^{\lg x-1}-13 \cdot 7^{\lg x-1}$. | ## Solution.
Domain: $x>0$.
From the condition
$$
7^{\lg x}-5 \cdot 5^{\lg x}=\frac{3}{5} \cdot 5^{\lg x}-\frac{13}{7} \cdot 7^{\lg x}, \quad 35 \cdot 7^{\lg x}+65 \cdot 7^{\lg x}=21 \cdot 5^{\lg x}+175 \cdot 5^{\lg x},
$$
$100 \cdot 7^{\lg x}=196 \cdot 5^{\lg x},\left(\frac{7}{5}\right)^{\lg x}=\left(\frac{7}{5}\r... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,828 |
7.057. $5^{x+6}-3^{x+7}=43 \cdot 5^{x+4}-19 \cdot 3^{x+5}$. | Solution.
We have $5^{6} \cdot 5^{x}-43 \cdot 5^{4} \cdot 5^{x}=3^{7} \cdot 3^{x}-19 \cdot 3^{5} \cdot 3^{x},\left(\frac{5}{3}\right)^{x}=\left(\frac{5}{3}\right)^{-3}$, from which $x=-3$.
Answer: -3. | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,829 |
7.058. $\frac{\log _{5}(\sqrt{2 x-7}+1)}{\log _{5}(\sqrt{2 x-7}+7)}=0.5$. | ## Solution.
Domain of definition: $2 x-7 \geq 0, x \geq \frac{7}{2}$.
From the condition
$\log _{5}(\sqrt{2 x-7}+1)=\frac{1}{2} \log _{5}(\sqrt{2 x-7}+7) \quad \log _{5}(\sqrt{2 x-7}+1)=\log _{5} \sqrt{\sqrt{2 x-7}+7}$, hence
$\sqrt{2 x-7}+1=\sqrt{\sqrt{2 x-7}+7} \Rightarrow(\sqrt{2 x-7})^{2}+2 \sqrt{2 x-7}+1=\sqr... | 5.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,830 |
7.060. $\sqrt{2^{x} \cdot \sqrt[3]{4^{x} \cdot 0.125^{1 / x}}}=4 \sqrt[3]{2}$. | Solution.
Domain of definition: $x \neq 0$.
Rewrite the equation as
$$
2^{\frac{x}{2}} \cdot 2^{\frac{x}{3}} \cdot 2^{-\frac{1}{2 x}}=2^{2} \cdot 2^{\frac{1}{3}}, 2^{\frac{x}{2}+\frac{x}{3}-\frac{1}{2 x}}=2^{2+\frac{1}{3}}
$$
from which
$\frac{x}{2}+\frac{x}{3}-\frac{1}{2 x}=\frac{7}{3}, 5 x^{2}-14 x-3=0$.
Then $... | -\frac{1}{5};3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,831 |
7.061. $\sqrt{2} \cdot 0.5^{\frac{5}{5^{x}+10}}-16^{\left.\frac{1}{2(\sqrt{x}+1}\right)}=0$. | ## Solution.
Domain of definition: $x \geq 0$.
From the condition
$$
2^{\frac{1}{2}} \cdot 2^{-\frac{5}{4 \sqrt{x}+10}}=2^{\frac{2}{\sqrt{x}+1}}, 2^{\frac{1}{2}-\frac{5}{4 \sqrt{x}+10}}=2^{\frac{2}{\sqrt{x}+1}}
$$
it follows that
$$
\frac{1}{2}-\frac{5}{4 \sqrt{x}+10}=\frac{2}{\sqrt{x}+1} \Rightarrow(\sqrt{x})^{2}... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,832 |
7.062. $8^{\frac{x-3}{3 x-7}} \sqrt[3]{\sqrt{0.25^{\frac{3 x-1}{x-1}}}}=1$. | Solution.
Domain of definition $\left\{\begin{array}{l}x \neq 1, \\ x \neq \frac{7}{3} .\end{array}\right.$
Rewrite the equation as
$$
2^{\frac{3 x-9}{3 x-7}} \cdot 2^{-\frac{3 x-1}{3 x-3}}=2^{0}, 2^{\frac{3 x-9}{3 x-7}-\frac{3 x-1}{3 x-3}}=2^{0}
$$
from which
$$
\frac{3 x-9}{3 x-7}-\frac{3 x-1}{3 x-3}=0 \Rightarr... | \frac{5}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,833 |
7.064. $0.6^{x}\left(\frac{25}{9}\right)^{x^{2}-12}=\left(\frac{27}{125}\right)^{3}$. | Solution.
We have
$$
\begin{aligned}
& \left(\frac{3}{5}\right)^{x}\left(\frac{3}{5}\right)^{-2 x^{2}+24}=\left(\frac{3}{5}\right)^{9},\left(\frac{3}{5}\right)^{-2 x^{2}+x+24}=\left(\frac{3}{5}\right)^{9} \\
& -2 x^{2}+x+24=9 ; 2 x^{2}-x-15=0
\end{aligned}
$$
from which $x_{1}=-\frac{5}{2}, x_{2}=3$.
Answer: $-\fra... | -\frac{5}{2};3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,834 |
7.066. $2^{\frac{1}{\sqrt{x}-1}} \cdot 0.5^{\frac{1}{\sqrt{x}+1}}=4^{\frac{\sqrt{x}}{x+\sqrt{x}}}$ | Solution.
Domain of definition: $0<x \neq 1$.
We have: $2^{\frac{1}{\sqrt{x}-1}} \cdot 2^{-\frac{1}{\sqrt{x}+1}}=2^{\frac{2 \sqrt{x}}{x+\sqrt{x}}}, 2^{\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}}=2^{\frac{2 \sqrt{x}}{x+\sqrt{x}}}$.
Then $\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}=\frac{2 \sqrt{x}}{x+\sqrt{x}}, x-\sqrt{... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,835 |
7.067. $2,5^{\frac{4+\sqrt{9-x}}{\sqrt{9-x}}} \cdot 0,4^{1-\sqrt{9-x}}=5^{10} \cdot 0,1^{5}$. | ## Solution.
Domain of definition: $9-x>0, x<9$.
Rewrite the equation as
$$
\left(\frac{5}{2}\right)^{\frac{4+\sqrt{9-x}}{\sqrt{9-x}}} \cdot\left(\frac{5}{2}\right)^{\sqrt{9-x}-1}=\left(\frac{5}{2}\right)^{5},\left(\frac{5}{2}\right)^{\frac{4+\sqrt{9-x}}{\sqrt{9-x}}+\sqrt{9-x}-1}=\left(\frac{5}{2}\right)^{5}
$$
The... | -7,8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,836 |
7.068. $2^{x^{2}-1}-3^{x^{2}}=3^{x^{2}-1}-2^{x^{2}+2}$. | ## Solution.
We have $\frac{2^{x^{2}}}{2}+4 \cdot 2^{x^{2}}=\frac{3^{x^{2}}}{3}+3^{x^{2}}, \frac{9}{2} \cdot 2^{x^{2}}=\frac{4}{3} \cdot 3^{x^{2}},\left(\frac{2}{3}\right)^{x^{2}}=\left(\frac{2}{3}\right)^{3}$.
Then $x^{2}=3$, from which $x_{1}=-\sqrt{3}, x_{2}=\sqrt{3}$.
Answer: $-\sqrt{3} ; \sqrt{3}$. | -\sqrt{3};\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,837 |
7.069. $\log _{\sqrt{5}}\left(4^{x}-6\right)-\log _{\sqrt{5}}\left(2^{x}-2\right)=2$. | Solution.
Domain of Definition (DOD): $\left\{\begin{array}{l}4^{x}-6>0 \\ 2^{x}-2>0 .\end{array}\right.$
We have $\log _{\sqrt{5}} \frac{4^{x}-6}{2^{x}-2}=2, \frac{2^{2 x}-6}{2^{2}-2}=5,2^{2 x}-5 \cdot 2^{x}+4=0$. Solving this equation as a quadratic in terms of $2^{x}$, we find $\left(2^{x}\right)=1$, from which we... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,838 |
7.070. $4^{\log _{9} x^{2}}+\log _{\sqrt{3}} 3=0.2\left(4^{2+\log _{9} x}-4^{\log _{9} x}\right)$. | ## Solution.
Domain of definition: $x>0$.
Rewrite the equation as
$4^{2 \log _{9} x}+2 \log _{3} 3=0.2\left(16 \cdot 4^{\log _{9} x}-4^{\log _{9} x}\right) \quad 4^{2 \log _{9} x}-3 \cdot 4^{\log _{9} x}+2=0$.
Solving this equation as a quadratic equation in terms of $4^{\log _{9} x}$, we find $\left(4^{\log _{9} x... | 1;3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,839 |
7.071. $3 \cdot 5^{2 x-1}-2 \cdot 5^{x-1}=0.2$. | ## Solution.
From the condition $3 \cdot 5^{2 x}-2 \cdot 5^{x}=1,3 \cdot 5^{2 x}-2 \cdot 5^{x}-1=0$. Solving this equation as a quadratic in terms of $5^{x}$, we get $5^{x}=-\frac{1}{3}, \varnothing$; or $5^{x}=1$, from which $x=0$.
Answer: 0. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,840 |
7.072. $10^{\frac{2}{x}}+25^{\frac{1}{x}}=4.25 \cdot 50^{\frac{1}{x}}$. | Solution.
Domain of definition: $x \neq 0$.
Dividing both sides of the equation by $25^{\frac{1}{x}}$, we get $2^{\frac{2}{x}}-4.25\left(2^{\frac{1}{x}}\right)+1=0$, from which, solving the equation as a quadratic in terms of $2^{\frac{1}{x}}$, we obtain $\left(2^{\frac{1}{x}}\right)=\frac{1}{4}$, hence $\left(\frac{... | x_{1}=-\frac{1}{2};x_{2}=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,841 |
7.073. $9^{x^{2}-1}-36 \cdot 3^{x^{2}-3}+3=0$. | Solution.
We have $\frac{9^{x^{2}}}{9}-36 \cdot \frac{3^{x^{2}}}{27}+3=0, \quad 3^{2 x^{2}}-12 \cdot 3^{x^{2}}+27=0$. Solving this
equation as a quadratic in terms of $3^{x^{2}}$, we get $3^{x^{2}}=3$, from which $x^{2}=1, x_{1,2}= \pm 1$, or $3^{x^{2}}=9$, from which $x^{2}=2, x_{3,4}= \pm \sqrt{2}$.
Answer: $-\sqrt... | -\sqrt{2};-1;1;\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,842 |
7.076. $9^{\sqrt{x-5}}-27=6 \cdot 3^{\sqrt{x-5}}$.
7.076. $9^{\sqrt{x-5}}-27=6 \cdot 3^{\sqrt{x-5}}$. | Solution.
Domain of definition: $x-5 \geq 0, x \geq 5$.
$$
3^{2 \sqrt{x-5}}-6 \cdot 3^{\sqrt{x-5}}-27=0
$$
We solve the equation as a quadratic equation in terms of $3^{\sqrt{x-5}}$. We have $3^{\sqrt{x-5}}=-3$. (not suitable) or $3^{\sqrt{x-5}}=9$, from which $\sqrt{x-5}=2$, or $x-5=4$. Then $x=9$.
Answer: 9. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,844 |
7.077. $17 \cdot 2^{\sqrt{x^{2}-8 x}}-8=2 \cdot 4^{\sqrt{x^{2}-8 x}}$. | Solution.
Domain of definition: $x^{2}-8 x \geq 0, x \in(-\infty ; 0] \cup[8 ;+\infty)$
We have $2 \cdot 2^{2 \sqrt{x^{2}-8 x}}-17 \cdot 2^{\sqrt{x^{2}-8 x}}+8=0$. Solving this equation as a quadratic equation in terms of $2^{\sqrt{x^{2}-8 x}}$, we get $2^{\sqrt{x^{2}-8 x}}=2^{-1}$, from which $\sqrt{x^{2}-8}= -1, \v... | -1;9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,845 |
7.078. $8^{\frac{2}{x}}-2^{\frac{3 x+3}{x}}+12=0$.
7.078. $8^{\frac{2}{x}}-2^{\frac{3 x+3}{x}}+12=0$. | Solution.
Domain of definition: $x \neq 0$.
Rewrite the equation as
$$
2^{\frac{6}{x}}-2^{3+\frac{3}{x}}+12=0,\left(2^{\frac{3}{x}}\right)^{2}-8 \cdot 2^{\frac{3}{x}}+12=0
$$
Solving this equation as a quadratic in terms of $2^{\frac{3}{x}}$, we get
$$
\begin{aligned}
& \left(2^{\frac{3}{x}}\right)_{1}=2, \text { ... | 3;\log_{6}8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,846 |
7.079. $2 \log _{x} 27-3 \log _{27} x=1$. | Solution.
Domain of definition: $0<x \neq 1$.
Let's switch to base 27. We have
$$
\frac{2}{\log _{27} x}-3 \log _{27} x-1=0 \Rightarrow 3 \log _{27}^{2} x+\log _{27} x-2=0
$$
Solving this equation as a quadratic equation in terms of $\log _{27} x$, we get $\left(\log _{27} x\right)_{1}=-1$, from which $x_{1}=\frac{... | \frac{1}{27};9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,847 |
7.080. $\lg (\sqrt{6+x}+6)=\frac{2}{\log _{\sqrt{x}} 10}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}6+x \geq 0, \\ x>0, \\ x \neq 1,\end{array} \quad 0<x \neq 1\right.$.
We will switch to base 10. We have
$\lg (\sqrt{6+x}+6)=2 \lg \sqrt{x}, \quad \lg (\sqrt{6+x}+6)=\lg x$.
Then $\sqrt{6+x}+6=x, \sqrt{6+x}=x-6 \Rightarrow\left\{\begin{array}{l}x^{2}-13 x+30=0... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,848 |
7.081. $\log _{5} x+\log _{x} 25=\operatorname{ctg}^{2} \frac{25 \pi}{6}$. | ## Solution.
Domain of definition: $0<x \neq 1$.
Switch to base 5. We have
$$
\begin{aligned}
& \log _{5} x+\frac{2}{\log _{5} x}=\left(\operatorname{ctg}\left(4 \pi+\frac{\pi}{6}\right)\right)^{2}, \log _{5} x+\frac{2}{\log _{5} x}=3 \Rightarrow \\
& \Rightarrow \log _{5}^{2} x-3 \log _{5} x+2=0
\end{aligned}
$$
S... | 5;25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,849 |
7.082. $x^{\frac{\lg x+5}{3}}=10^{5+\lg x}$. | Solution.
Domain of definition: $0<x \neq 1$.
Taking the logarithm of both sides of the equation with base 10, we have
$$
\lg x^{\frac{\lg x+5}{3}}=\lg 10^{5+\lg x}, \frac{\lg x+5}{3} \lg x=(5+\lg x) \lg 10, \lg ^{2} x+2 \lg x-15=0
$$
Solving this equation as a quadratic equation in terms of $\lg x$, we get $(\lg x... | 10^{-5};10^{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,850 |
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