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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](http://meta.math.stackexchange.com/questions/5020/tex-latex-mathjax-basic-tutorial-and-quick-reference), [here](http://meta.stackexchange.com/questions/68388/there-should-be-universal-latex-mathjax-guide-for-sites-supporting-it/70559#70559), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latex).2012-09-26
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    i copied and pasted what you posted. it didnt work. BTW that is my question2012-09-26
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    Did you also copy the dollar signs, and did *not* prefix it with four spaces?2012-09-26
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    Try $a = 3\sqrt[3]{a}$ and cube both sides.2018-07-28

4 Answers 4

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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!2012-09-26
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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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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.2012-09-26
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    @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.2012-09-26
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    i posted it in the original question man, I only use my calculator to check..2012-09-26
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    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.2012-09-26
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    but how do i cube root logx^(27)? Thats the part im stuck with.2012-09-26
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    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.]2012-09-26
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    Ok so using that, I got x = 10^(sqrt(27)) which is right but, I never got 1, which is also a solution?2012-09-26
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    @vincentbelkin: In that case my guess is that you did some dividing, rather than factorising. You should get to $u^3=27u$, right? There are $3$ possible values $u$ can take. To find them all, factorise the expression.2012-09-26
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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.2012-09-26
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    Posted an answer man! not sure if its right though2012-09-26