# How do you solve  5^(x - 1) = 3^x?

Mar 16, 2016

$x \approx 3.15$

#### Explanation:

$1$. Since the left and right sides of the equation do not have the same base, start by taking the log of both sides.

${5}^{x - 1} = {3}^{x}$

$\log \left({5}^{x - 1}\right) = \log \left({3}^{x}\right)$

$2$. Use the log property, ${\log}_{\textcolor{p u r p \le}{b}} \left({\textcolor{red}{m}}^{\textcolor{b l u e}{n}}\right) = \textcolor{b l u e}{n} \cdot {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right)$, to simplify both sides of the equation.

$\left(x - 1\right) \log 5 = x \log 3$

$3$. Expand the brackets.

$x \log 5 - \log 5 = x \log 3$

$4$. Group all like terms together such that the terms with the variable, $x$, are on the left side and $\log 5$ is on the right side.

$x \log 5 - x \log 3 = \log 5$

$5$. Factor out $x$ from the terms on the left side of the equation.

$x \left(\log 5 - \log 3\right) = \log 5$

$6$. Solve for $x$.

$x = \log \frac{5}{\log 5 - \log 3}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x \approx 3.15 \textcolor{w h i t e}{\frac{a}{a}} |}}}$