# How to solve the following equation? log_3(log_x(log_4 16))=−1.

How to solve the following equation? ${\mathrm{log}}_{3}\left({\mathrm{log}}_{x}\left({\mathrm{log}}_{4}16\right)\right)=-1$
I am trying to solve this equation for $x$. This is what I have so far:
${\mathrm{log}}_{3}\left({\mathrm{log}}_{x}2\right)=-1.$
Okay, now I have this:
$\mathrm{log}2=\left(1/3\right)\mathrm{log}x$
How do I isolate x from here?
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Aidyn Mccarthy
Adriano's method is another method, I will continue where you stopped:
${\mathrm{log}}_{3}\left({\mathrm{log}}_{x}2\right)=-1⇔\frac{1}{3}={\mathrm{log}}_{x}\left(2\right)$
by the definition of the logarithm.
Now ${x}^{1/3}=2⇔x={2}^{3}=8$, again by the definition of the logarithm.
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Danika Mckay
Hint: There is no multiplication going on anywhere in this equation (just nested logs). Try thinking of the stuff inside the brackets as a single unit and convert to exponential form, working your way from the outside to the inside. If it helps, use substitutions. For example, let $u={\mathrm{log}}_{x}\left({\mathrm{log}}_{4}16\right)\right)$, so that we have:
${\mathrm{log}}_{3}u=-1$
Converting to exponential form, we have:
$u={3}^{-1}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{x}\left({\mathrm{log}}_{4}16\right)=\frac{1}{3}$
Now repeat.
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