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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. – Brian M. Scott Apr 29 '12 at 23:45
You could've checked that $\log e^e = e\log e=e$, and then $\log e =1$, which is what you want. – Pedro Tamaroff Apr 29 '12 at 23:47
Cool thanks, I didn't see that connection. – Kurt Apr 29 '12 at 23:48
up vote 3 down vote accepted

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 $b\in \mathbb{R}-\mathbb{Q}$ – Abdelmajid Khadari Apr 30 '12 at 0:46
@AbdelmajidKhadari Yes. It is a direct consequence of the definition of the logarithm. – Pedro Tamaroff Jul 6 '12 at 6:12

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