# Log and Natural Log Question [closed]

Prove $x^{\ln(x)} = e^{(\ln(x))^3}$

($x$ raised to $\ln(x)$ $=$ $e$ raised to ($\ln(x)$ cubed).

I have an exam in 20 minutes. Appreciate the answers.

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## closed as off-topic by user127.0.0.1, user91500, Davide Giraudo, Avitus, Umberto P.Apr 7 at 13:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user127.0.0.1, user91500, Davide Giraudo, Avitus, Umberto P.
If this question can be reworded to fit the rules in the help center, please edit the question.

As usual, it may be the question is poorly posted. It may have been "solve the equation $\;x^{\log x}=e^{\log^3x}\;$" , and then

$$x^{\log x}=e^{\log^2x}\implies x^{\log x}=e^{\log^3x}\iff \log^2x=\log^3x\iff$$

$$\log^2x(\log x-1)=0\iff\begin{cases}\log x=0\iff x=1\\\text{or}\\\log x=1\iff x=e\end{cases}$$

Oh, well: it is now more than 20 minutes after the OP asked and the exam is over almost surely...

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Perhaps I'm mistaken but I believe it should be squared :$x^{\ln(x)}=(e^{\ln(x)})^{\ln(x)}=e^{(\ln(x))^{2}}$.

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No, it isn't that. The question is correct –  TheEconomist Jan 19 at 5:12
@TheEconomist Apparently not. –  T. Bongers Jan 19 at 5:14
If the question's correct then it is a rather pretty easy one. It's answer is: the equality is wrong. –  DonAntonio Jan 19 at 9:51
Why does this question deserve 5 downvotes? The question was correct. –  TheEconomist Jan 21 at 12:20
@TheEconomist, if the OP was correct, the only possible answer is: the equality is wrong...but you did not asked whether the equality is right or wrong, but to prove that equality, which of course cannot be done. –  DonAntonio Jan 21 at 15:46

There is nothing to prove the two expressions are not equivilent, this can be seen by taking the logarithm of both sides.

$$\ln(x)+\ln(\ln(x))\ne \ln(x)^3$$

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Taking the logarithm on both side you should find that $\text{ln}(x)^2 = \text{ln}(x)^3$ , which has two solutions. –  Olivier Jan 19 at 5:44
@Olivier User71352 changed the question, look at the un-edited version –  Ethan Jan 19 at 5:46
For any $a$
$$x^a = e^{a \ln(x)}$$ Put $a = \ln(x)$ and you get $$x^{\ln(x)} = e^{\ln(x) \cdot \ln(x)} = e^{\ln(x)^2}$$
$$3^{\ln(3)} \approx 3.3433 \\ e^{\ln(3)^2} \approx 3.3433$$