# How do you solve 10^(3x)+7=19?

Aug 24, 2016

$x = \frac{\log 12}{3}$

#### Explanation:

The first thing we do is subtract 7 from both sides:

${10}^{3 x} = 12$

Now we ad logarithms to both sides:

$\log {10}^{3 x} = \log 12$

One of the properties of logarithms is that we can pull the exponent to the front, like this:

$3 x \cdot \log 10 = \log 12$

Whenever you see a $\log$ without a base it is base 10 and one of the properties of logarithms is that any $\log$ of a number equal to its base is equal to 1.
${\log}_{x} x = 1$

So ${\log}_{10} 10 = 1$
leaving us with:

$3 x = \log 12$

so

$x = \frac{\log 12}{3}$