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$$\ln2e^{2x}$$

Here are the two results I came up with: $$2x(\ln2e)$$ $$2x(\ln2+\ln e)$$ $$2x(\ln2 + 1)$$ $$2x\ln2+2x$$

and

$$\ln2+\ln e^{2x}$$ $$\ln2+2x\ln e$$

I am sort of leaning towards the first result I got but I am not really sure. Could someone explain whether or not it is correct? I have looked at the log rules but I cannot recall which ones have priority over others.

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The first one isn't correct because it is $\ln(2e^{2x})$, not $\ln((2e)^{2x})$. The $2$ hasn't been raised to the power of $2x$. –  Henry T. Horton Apr 26 '13 at 2:08
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3 Answers

up vote 3 down vote accepted

The second result is correct. The power of 2x is not applied to the entire argument of the logarithm and thus cannot be factored out of the logarithm, so the first result is incorrect.

Your second result ln 2 + 2x ln e can be simplified further to be ln 2 + 2x since ln e = 1.

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None of the identities have priority over one another. log(x) is a function with identities, not a system of operations. There's no way two results of the function could not equal themselves.

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More generally, $\ \log a +n \log b = \log (ab^n) \neq \ n \ \log (ab) = \log (a^n b^n) = \log (ab)^n$

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