# How do you solve 2^x = 3^(x-1) ?

Dec 11, 2015

$x = {\log}_{\frac{2}{3}} 3$

#### Explanation:

$\textcolor{w h i t e}{\times x} {2}^{x} = {3}^{x - 1}$

$\implies {\log}_{3} {2}^{x} = x - 1$

The logarithm of the ${x}^{t h}$ power of a number is $x$ times the logarithm of the number itself:
$\textcolor{w h i t e}{\times x} x {\log}_{3} 2 = x - 1$

Multiply both sides by $\textcolor{red}{\frac{1}{x}}$
$\textcolor{w h i t e}{\times x} \textcolor{red}{\frac{1}{x} \cdot} x {\log}_{3} 2 = \textcolor{red}{\frac{1}{x} \cdot} \left(x - 1\right)$
$\implies \frac{x - 1}{x} = {\log}_{3} 2$

Add $\textcolor{red}{- 1}$ to both sides:
$\textcolor{w h i t e}{\times x} \frac{x - 1}{x} \textcolor{red}{- 1} = {\log}_{3} 2 \textcolor{red}{- 1}$
$\implies \frac{1}{x} = {\log}_{3} 2 - 1$
$\implies \frac{1}{x} = {\log}_{3} 2 - \textcolor{b l u e}{{\log}_{3} 3} \textcolor{w h i t e}{\times \times x}$ (because for $\forall a \in \mathbb{R} , {a}^{1} = a$)

The logarithm of the ratio of two numbers is the difference of the logarithms:
$\implies \frac{1}{x} = {\log}_{3} \left(\frac{2}{3}\right)$
$\implies x = {\log}_{\frac{2}{3}} 3$