# How do I find log_27 9?

Mar 15, 2015

The answer is $\frac{2}{3}$.

Method 1
${\log}_{27} \left(9\right)$ can be interpreted as "$27$ to what power is equal to $9$.

Since ${27}^{\frac{2}{3}} = {3}^{2} = 9 , {\log}_{27} \left(9\right) = \frac{2}{3}$.

Method 2
If it is difficult to see directly, another technique is to change the base:

${\log}_{27} \left(9\right)$ can be rewritten as $\frac{{\log}_{3} \left(9\right)}{\log} _ 3 \left(27\right) = \frac{2}{3}$

Mar 15, 2015

As has already been pointed out, ${\log}_{27} \left(9\right)$ is the exponent needed on 27 to get 9. It is the solution to ${27}^{x} = 9$.

${27}^{x} = 9$

Because $27 = {3}^{3}$ and $9 = {3}^{2}$, we must want

${\left({3}^{3}\right)}^{x} = {3}^{2}$

But ${\left({3}^{3}\right)}^{x} = {3}^{3 x}$, so we also want:

${3}^{3 x} = {3}^{2}$.

Now, it is a property of exponents the: if $3$ to this power equals $3$ to that power, then this power must be equal to that power.

So we need $3 x = 2$, which means we need #x=2/3.

Check: ${27}^{\frac{2}{3}} = {\sqrt[3]{27}}^{2} = {3}^{2} = 9$