# Logarithms and Exponents

## Exponent Laws
- $a^m \cdot a^n = a^{m+n}$
- $\frac{a^m}{a^n} = a^{m-n}$
- $(a^m)^n = a^{mn}$
- $(ab)^n = a^n b^n$
- $a^0 = 1$ for $a \neq 0$
- $a^{-n} = \frac{1}{a^n}$
- $a^{1/n} = \sqrt[n]{a}$

## Logarithm Definition
$\log_a x = y \iff a^y = x$ where $a > 0, a \neq 1, x > 0$

## Logarithm Properties
- $\log_a(xy) = \log_a x + \log_a y$
- $\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y$
- $\log_a(x^n) = n \log_a x$
- $\log_a a = 1$
- $\log_a 1 = 0$

## Change of Base
$\log_a x = \frac{\log_b x}{\log_b a} = \frac{\ln x}{\ln a}$
$\log_a b = \frac{1}{\log_b a}$
$\log_a b \cdot \log_b c = \log_a c$

## Common Logarithm Identities
- $a^{\log_a x} = x$
- $\log_{a^k} x = \frac{1}{k} \log_a x$
- $\log_a x = \log_a y \implies x = y$

## JEE Tips
- Domain of $\log_a f(x)$: need $f(x) > 0$
- $\log_a x > \log_a y$: if $a > 1$ then $x > y$; if $0 < a < 1$ then $x < y$
- When solving $\log$ equations, always verify solutions satisfy domain constraints