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$$\log_2\left\{\log_3\left[\log_4\left(x^{3x}\right)\right]\right\} = 0$$

How would I go about solving this? I tried doing $\log_4(x^{3x}))=0$ but I don't know how to incorporate the other logs

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    $\begingroup$ Try peeling it like an onion. If $\log_2(\text{something})=0$, can you tell what the value of "something" is $\endgroup$ – Dilip Sarwate Sep 22 '14 at 20:07
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If $\log_2(\text{soemthing})=0$, then $\text{something}=1$.

So $\log_3 \log_4\left(x^{3x}\right)=1$.

If $\log_3(\text{something})=1$, then $\text{something}=3$.

So $\log_4(x^{3x})=3.$

Now recall that $\log_4(x^{3x})=3x\log_4 x$, so we get $3x\log_4 x = 3$.

It follows that $x\log_4 x = 1$. That implies $\log_4 (x^x)=1$, so $x^x=4$.

Doubtless you know that $2^2=4$, so $x=2$ is a solution. If $x>2$ then $x^x>4$. If $x<2$ then $x^x<4$, although that last fact is a bit more involved than it might superficially look.

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Work from the outside in. Recall that: $$ \log_b(y) = a \iff b^a = y $$ Hence, we get: \begin{align*} 2^0 &= \log_3(\log_4(x^{3x})) \\ 3^1 &= \log_4(x^{3x}) \\ 4^3 &= x^{3x} \\ \end{align*} Now notice that $4^3 = (2^2)^3 = 2^6$. By inspection, we see that $x = 2$ is one possible solution.

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