Help with Evaluating a Logarithm A precalculus text asks us to evaluate $\log_{8}\dfrac{\sqrt{2}\cdot\sqrt[3]{256}}{\sqrt[6]{32}}$
I do the following:
$\log_{8}\dfrac{\sqrt{2}\cdot\sqrt[3]{(2^2)^3\cdot 2^2}}{\sqrt[6]{2^3\cdot 2^2}}$
$\equiv \log_{8}\dfrac{\sqrt{2}\cdot 2^2\cdot\sqrt[3]{2^2}}{\sqrt{2}\cdot\sqrt[6]{2^2}}$  
$\equiv \log_{8}\dfrac{2^2\cdot\sqrt[3]{2^2}}{\sqrt[3]{2}}$
and then I'm stumped.
Hints?

$$ \log_{8}\left(\frac{2^2\cdot\sqrt[3]{2^2}}{\sqrt[3]{2}}\right) = \log_8 \left(2^2\cdot \sqrt[3]2\right) = \log_8 2^{7/3} = \frac 73 \log_8(2)= \frac 73 \cdot \frac 13 = \frac 79$$
$$8^{1/3} = 2 \iff \log_8 2 = \frac 13$$
