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I got to this: $2\log_{10}x^{\frac{1}{2}}+3\log_{10}x^{\frac{1}{3}}$.

Now, usually you bring the exponent the the front and that would yield:


And that's it?

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Check your arithmetic: $\frac12(2)\ne 2$. – Brian M. Scott Jul 15 '12 at 21:06
up vote 3 down vote accepted

$$ \frac{1}{2}(2)\log_{10}x+\frac{1}{3}(3)\log_{10}x \neq 2\log_{10}x+3\log_{10}x$$

$1/2 * 2 = 1$ so do $1/3 * 3.$

So the correct thing is $$\frac{1}{2}(2)\log_{10}x+\frac{1}{3}(3)\log_{10}x = \log_{10}x+\log_{10}x = 2\log_{10}x$$ Bring the exponent inside the $\log$ if you like.

The other way around: bring $1/2$ into $\log_{10} \sqrt{x}$ to become $\log_{10} \sqrt{x}^2 = ??.$

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