# How do you solve 5^ { x + 2} = 5^ { 9} ?

May 6, 2017

$x = 7$

#### Explanation:

When the base number on the LEFT-hand side equals to the base number on the RIGHT-hand side, we can simply equate their exponents/power.

${5}^{x + 2} = {5}^{9}$

By equating the exponents on the LHS and RHS, we get:

$x + 2 = 9$
$\textcolor{red}{x = 7}$

Double checking the solution,

${5}^{7 + 2} = {5}^{9}$
${5}^{9} = {5}^{9}$ Hurray!

Alternatively , we can use the logarithm to solve.
By adding ${\log}_{5}$ to both sides, we get:

${\log}_{5} {5}^{x + 2} = {\log}_{5} {5}^{9}$

The special properties of logarithmic functions allow us to "bring" down the exponent as such:

$\left(x + 2\right) {\log}_{5} 5 = 9 {\log}_{5} 5$

When the base b, of the logarithmic functions is the same as the number, ${\log}_{b} b = 1$.

Therefore, $\left(x + 2\right)$ x $1 = 9$ x $1$

$x + 2 = 9$
$\textcolor{red}{x = 7}$