Question

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

Answer 1

`x=(log12)/3`

Explanation:

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

`10^(3x)=12`

Now we ad logarithms to both sides:

`log 10^(3x)=log12`

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

`3x*log10=log12`

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:

`3x=log 12`

so

`x=(log 12)/3`