Bodonimhk

2022-10-14

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

amilazamiyn

Expert

${\mathrm{log}}_{8}\left(\frac{{2}^{2}\cdot \sqrt[3]{{2}^{2}}}{\sqrt[3]{2}}\right)={\mathrm{log}}_{8}\left({2}^{2}\cdot \sqrt[3]{2}\right)={\mathrm{log}}_{8}{2}^{7/3}=\frac{7}{3}{\mathrm{log}}_{8}\left(2\right)=\frac{7}{3}\cdot \frac{1}{3}=\frac{7}{9}$
${8}^{1/3}=2\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{8}2=\frac{1}{3}$

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