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How does $\lg{x}-\lg{\sqrt[3]{x}}-\lg{\sqrt[6]{x}}$ simplify to $\lg{\sqrt{x}}$?

I've tried to get Bagatrix Algebra Solved! to solve it, but it even got the wrong answer... (I checked it by replacing x with 5 and typing it out on a calculator..)

No matter what I do, I end up with an answer that is correct and a bit simplified, but not as simplified as $\lg{\sqrt{x}}$.

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    hint $\log(x^a)=a\log(x)$2012-10-16
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    http://meta.math.stackexchange.com/questions/6299/should-the-algebra-tag-be-blacklisted?cb=12012-10-16
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    @AD. So when should I use the $\text{algebra}$ tag?2012-10-16
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    Never. For this kind of thing, you should use `algebra-precalculus`.2012-10-16
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    @StudentofHogwarts No one should in fact use that tag. Please read the tag info http://math.stackexchange.com/tags/algebra/info2012-10-17

4 Answers 4

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$$\begin{align*}\lg x-\lg\sqrt[3]{x}-\lg\sqrt[6]{x}&=\lg\frac{x}{x^{1/3}\cdot x^{1/6}}\\ &=\lg\frac{x}{x^{\frac13+\frac16}}\\ &=\lg\frac{x}{x^{1/2}}\\ &=\lg\left(x^{1-\frac12}\right)\\ &=\lg x^{1/2}\\ &=\lg\sqrt x\;. \end{align*}$$

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    How did you go from the second line to the third line?2012-10-16
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    @StudentofHogwarts $x^{1/3}\cdot x^{1/6} = x^{1/3 + 1/6} = x^{1/2}$.2012-10-16
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    @StudentofHogwarts: I’ve added a couple of extra steps that should answer the question.2012-10-16
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    @Brian Thanks for an excellent answer! It seems like I didn't find the answer because I didn't figure out that you could do like you did from the third line to the fourth line..2012-10-16
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Do you know these two identities?

$$\sqrt[n]{x}=x^{1/n}$$

$$\lg a^b=b\lg a$$

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    Yes -----------------2012-10-16
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Note that $$\lg\sqrt[n]{x}=\lg\left(x^{1/n}\right)=\frac1n\lg x$$ for any positive integer $n$. That should get you where you need to go, since $\sqrt{x}=\sqrt[2]{x}$.

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You probably want to assume $x > 0$ to have a canonical choice for the branches of $\sqrt[3]x$ and $\sqrt[6]x$. Hint: $\log x^n = n \log x$ generalises to rational $n$.