# How do you solve  5 times 6^x = 6 times 5^x?

May 14, 2016

$x = 1$

#### Explanation:

Given,

$5 \cdot {6}^{x} = 6 \cdot {5}^{x}$

Take the logarithm of both sides since the bases are not the same.

$\log \left(5 \cdot {6}^{x}\right) = \log \left(6 \cdot {5}^{x}\right)$

Using the logarithmic property, ${\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{b l u e}{x} \cdot \textcolor{red}{y}\right) = {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{b l u e}{x}\right) + {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{y}\right)$, the equation becomes,

$\log \left(5\right) + \log \left({6}^{x}\right) = \log \left(6\right) + \log \left({5}^{x}\right)$

Group all terms with $x$ on one side of the equation.

$\log \left({6}^{x}\right) - \log \left({5}^{x}\right) = \log \left(6\right) - \log \left(5\right)$

Using the logarithmic property, ${\log}_{\textcolor{p u r p \le}{b}} \left({\textcolor{b l u e}{x}}^{\textcolor{red}{y}}\right) = \textcolor{red}{y} \cdot {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{b l u e}{x}\right)$, the equation becomes,

$x \log \left(6\right) - x \log \left(5\right) = \log \left(6\right) - \log \left(5\right)$

Factor out $x$ from the left side.

$x \left(\log \left(6\right) - \log \left(5\right)\right) = \log \left(6\right) - \log \left(5\right)$

Solve for $x$.

$x = \frac{\log \left(6\right) - \log \left(5\right)}{\log \left(6\right) - \log \left(5\right)}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x = 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$