# How do I cube/square a logarithm?

Btw, please don't give me the answer. I just wanna know how to raise a logarithm to its cube cause I'm stuck in this part, but don't solve it for me.

$$\log \sqrt[3]x = \sqrt[3]{\log x}$$ I tried different ways but. When I input the values in my calculator, it just doesn't match.

My answer: is it right? $$1/3\log x =\sqrt[3]{\log x}$$ $$\log x = 3\sqrt[3]{\log x}$$ $$a = 3\sqrt[3]{\log x}$$ $$a^3 = \log x^{27}$$ $$\log x^3 = \log x^{27}$$ $$\log x^{24} = 0$$ $$x = \sqrt[24]{1}$$ $$x = 1$$

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Did you want to write something like $\log\sqrt[3]{x}=\sqrt[3]{\log x}$? This can be obtained as $\log\sqrt[3]{x}=\sqrt[3]{\log x}$. For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Sep 26 '12 at 11:58
i copied and pasted what you posted. it didnt work. BTW that is my question – Maximus Programus Sep 26 '12 at 11:59
Did you also copy the dollar signs, and did not prefix it with four spaces? – celtschk Sep 26 '12 at 12:08

You can write $\log \sqrt[3] x$ in terms of $\log x$ using the laws of logarithms. So substitute $u=\log x$ and solve for $u$ (after rearranging you have a cubic equation in $u$) then substitute back and solve for $x$.

Edit: Your error in your working is thinking that $a^3 = \log x^3$. In fact, $a^3 = (\log x)^3$.

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Tried it man! not sure if I did it correctly though.. the calculator says its right. But both of the sides are equal to zero so its kinda fishy. – Maximus Programus Sep 26 '12 at 12:20
@vincentbelkin: Show your working if you're stuck. I think the problem here might be that you're using a calculator! EDIT: also see the edit to my post. – Clive Newstead Sep 26 '12 at 12:23
i posted it in the original question man, I only use my calculator to check.. – Maximus Programus Sep 26 '12 at 12:24
I edited my post. When you substitute $a=\log x$, substitute it throughout, not just for one instance of $\log x$. Then solve for $a$, and then sub back and solve for $x$; that way you will make fewer errors. In fact, $x=1$ is one solution, but there are more that you missed in error. – Clive Newstead Sep 26 '12 at 12:25
You're on the wrong track. Let me start you off. You know that $\log \sqrt[3]{x} = \frac{1}{3}\log x$, so the equation $\log \sqrt[3]{x} = \sqrt[3]{\log x}$ becomes $\frac{1}{3}u = \sqrt[3]{u}$ when we substitute $u=\log x$. Forget $x$ for now and just solve this for $u$. Then substitute back and solve for $x$. [You seem to be repeatedly making the error of thinking that $(\log x)^a = \log (x^a)$ - this is not the case.] – Clive Newstead Sep 26 '12 at 12:34

1) Remember that $\,\log x^n=n\log x\,$

2) Now just note that $\,\log \sqrt[3] x=\log x^{1/3}\,$

Anyways, it is not true in general that $\,\log \sqrt[3] x=\sqrt[3]{\log x}\,$

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$\log$ is not so friendly with square/cube roots.., i.e. your identity is not true in general. Are you looking for its solution $x$?

Do you know that $\sqrt[3]x=x^{\frac13}$?

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With odd roots it is just fine. Problems can arise with even roots if we're not sufficiently careful. – DonAntonio Sep 26 '12 at 12:20
Posted an answer man! not sure if its right though – Maximus Programus Sep 26 '12 at 12:23

One method you may attempt; when you get to the step:

$\log(x)^3=27\log(x)$

Let $u= \log(x)$

and solve cubic: $u^3 - 27u = 0$

so that: $u(u^2 - 27)$=0

therefore: $u(u-\sqrt{27})(u + \sqrt{27})= 0$

Now find the roots then resubstitute $\log(x) = u$; to get the answer

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seems a similar answer was already posted! – student101 Sep 26 '12 at 12:33