# How do you solve 12=10^(x+5)-7?

Sep 10, 2016

#### Answer:

$x = {\log}_{10} \left(19\right) - 5$

#### Explanation:

We have: $12 = {10}^{x + 5} - 7$

Let's add $7$ to both sides of the equation:

$\implies 19 = {10}^{x + 5}$

$\implies {10}^{x + 5} = 19$

Then, let's apply ${\log}_{10}$ to both sides of the equation:

$\implies {\log}_{10} \left({10}^{x + 5}\right) = {\log}_{10} \left(19\right)$

Using the laws of logarithms:

$\implies \left(x + 5\right) {\log}_{10} \left(10\right) = {\log}_{10} \left(19\right)$

$\implies \left(x + 5\right) \cdot 1 = {\log}_{10} \left(19\right)$

$\implies x + 5 = {\log}_{10} \left(19\right)$

Now, to solve for $x$, let's subtract $5$ from both sides:

$\implies x = {\log}_{10} \left(19\right) - 5$