# How do you solve  5^x = 28?

Mar 25, 2016

$x \approx 2.07$

#### Explanation:

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

${5}^{x} = 28$

$\log \left({5}^{x}\right) = \log \left(28\right)$

$2$. 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 the left side of the equation.

$x \log \left(5\right) = \log \left(28\right)$

$3$. Solve for $x$.

$x = \frac{\log \left(28\right)}{\log \left(5\right)}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x \approx 2.07 \textcolor{w h i t e}{\frac{a}{a}} |}}}$