Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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


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

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.

share|cite|improve this question
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
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.

share|cite|improve this answer

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.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.