# How do you solve −10 + log_ 3 (n + 3) = −10?

Jan 25, 2016

$n = - 2$

#### Explanation:

$- 10 + {\log}_{3} \left(n + 3\right) = - 10$

First of all, add $10$ on both sides of the equation:

$\iff \textcolor{w h i t e}{\times} {\log}_{3} \left(n + 3\right) = 0$

To "get rid" of the ${\log}_{3}$ term, you need to exponentiate the expression to the base $3$, since ${a}^{x}$ is the inverse function for ${\log}_{a} \left(x\right)$ and thus, both ${a}^{{\log}_{a} \left(x\right)} = x$ and ${\log}_{a} \left({a}^{x}\right) = x$ hold.

$\iff \textcolor{w h i t e}{\times} {3}^{{\log}_{3} \left(n + 3\right)} = {3}^{0}$

... don't forget that for any number $b$ you can compute ${b}^{0} = 1$

$\iff \textcolor{w h i t e}{\times \times x} n + 3 = 1$

... subtract $3$ on both sides of the equation...

$\iff \textcolor{w h i t e}{\times \times \times} n = - 2$