# How does this simplify to $\lg{\sqrt{x}}$?

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 $\log(x^a)=a\log(x)$ –  karakfa Oct 16 '12 at 21:17
@AD. So when should I use the $\text{algebra}$ tag? –  Student of Hogwarts Oct 16 '12 at 21:32
Never. For this kind of thing, you should use algebra-precalculus. –  Cameron Buie Oct 16 '12 at 22:13
@StudentofHogwarts No one should in fact use that tag. Please read the tag info math.stackexchange.com/tags/algebra/info –  AD. Oct 17 '12 at 5:09

\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? –  Student of Hogwarts Oct 16 '12 at 21:24
@StudentofHogwarts $x^{1/3}\cdot x^{1/6} = x^{1/3 + 1/6} = x^{1/2}$. –  MJD Oct 16 '12 at 21:25
@StudentofHogwarts: I’ve added a couple of extra steps that should answer the question. –  Brian M. Scott Oct 16 '12 at 21:28
@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.. –  Student of Hogwarts Oct 16 '12 at 21:30

Do you know these two identities?

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

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

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Yes ----------------- –  Student of Hogwarts Oct 16 '12 at 21:34

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 $n$.

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