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I worked out this question, and I wanted to see if my understanding of the concepts involved is sound.

Solve for $x$ $$\ln(\ln(x))=1$$

$$e^1=\ln(x)$$ $$e^e=x$$ Since any number raised to $1$ is just itself, the final answer could be expressed as $x=e^e$ This is not a homework question, but a question I stumbled across which I find interesting: I want to see if my understanding is correct.

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    Yes, this is correct.2012-04-29
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    You could've checked that $\log e^e = e\log e=e$, and then $\log e =1$, which is what you want.2012-04-29
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    Cool thanks, I didn't see that connection.2012-04-29

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It is correct. To verify, plug the answer back into the equation and check that both sides are equal:

\begin{align*} \ln(\ln(e^e)) = \ln(e \ln(e)) = \ln(e \cdot 1) = \ln(e) = 1 \end{align*}

Remember that $\ln(a^b)=b\ln(a)$ and $\ln(e) = 1$.

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    is that $\ln(a^b)=b\ln(a)$ correct !!? for $b\in \mathbb{R}-\mathbb{Q}$2012-04-30
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    @AbdelmajidKhadari Yes. It is a direct consequence of the definition of the logarithm.2012-07-06