# How do you solve log_3 3 + log_2 8 + log_4 64?

##### 1 Answer
Nov 19, 2015

I found $7$.

#### Explanation:

You can use the definition of log as:
${\log}_{b} x = a \to x = {b}^{a}$
So, basically you can substitute the exponents of the base that you need to get the argument:
${\log}_{3} \left(3\right) = 1$ because ${3}^{1} = 3$
${\log}_{2} \left(8\right) = 3$ because ${2}^{3} = 8$
${\log}_{4} \left(64\right) = 3$ because ${4}^{3} = 64$
And finally:
$1 + 3 + 3 = 7$