# How do you solve 2^x = 7?

Mar 16, 2016

$x \approx 2.81$

#### 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.

${2}^{x} = 7$

$\log \left({2}^{x}\right) = \log \left(7\right)$

$2$. 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 $\log \left({2}^{x}\right)$.

$x \log \left(2\right) = \log \left(7\right)$

$3$. Solve for $x$.

$x = \log \frac{7}{\log} \left(2\right)$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x \approx 2.81 \textcolor{w h i t e}{\frac{a}{a}} |}}}$