Question

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

Answer 1

`n = -2`

Explanation:

`-10 + log_3(n+3) = -10`

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

`<=> color(white)(xx) log_3(n+3) = 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(x)` and thus, both `a^(log_a(x)) = x` and `log_a(a^x) = x` hold.

`<=> color(white)(xx) 3^(log_3(n+3)) = 3^0`

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

`<=> color(white)(xxxxx) n + 3 = 1`

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

`<=> color(white)(xxxxxx) n = -2`