# How do you evaluate log_64 8?

Sep 18, 2016

${\log}_{64} 8 = \frac{1}{2}$.

#### Explanation:

We will use the Defn. of $\log$ fun. to find the desired value.

Recall that, ${\log}_{b} a = x \iff {b}^{x} = a$.

So, ${\log}_{64} 8 = x \Rightarrow {64}^{x} = 8 \Rightarrow {\left({8}^{2}\right)}^{x} = 8 \Rightarrow {8}^{2 x} = {8}^{1}$.

As $\log$ fun. is $1 - 1$, we must have, $2 x = 1 , \text{ giving, } x = \frac{1}{2}$.

$\therefore {\log}_{64} 8 = \frac{1}{2}$.

Sep 18, 2016

${\log}_{64} 8 = \frac{1}{2}$

#### Explanation:

When I write ${\log}_{a} b = c$, I ask to what power I raise the base $a$ to get $b$; the answer here is $c$, i.e. ${a}^{c} = b$. For ${\log}_{10} 10$, ${\log}_{10} 100$, ${\log}_{10} 1000$, ${\log}_{10} 1000000$, I can immediately write $1 , 2 , 3 , 6$ without thinking too much.

For your problem, ${\log}_{64} 8$, I look for the power to which I raise $64$ to get $8$. But of course, ${8}^{2} = 64$, or $\sqrt{64} = 8$, ${64}^{\frac{1}{2}} = 8$. So ${\log}_{64} 8 = \frac{1}{2}$.