Logarithm rules, which one has priority? \(\displaystyle{\ln{{2}}}{e}^{{{2}{x}}}\)

Tony Mccarthy

Tony Mccarthy

Answered question

2022-03-21

Logarithm rules, which one has priority? ln2e2x
ln2e2x
Here are the two results I came up with:
2x(ln2e)
2x(ln2+lne)
2x(ln2+1)
2xln2+2x
and
ln2+lne2x
ln2+2xlne
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.

Answer & Explanation

Theodore Davila

Theodore Davila

Beginner2022-03-22Added 14 answers

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 ln2+2xlne can be simplified further to be ln2+2x since lne=1.
Cecilia Nolan

Cecilia Nolan

Beginner2022-03-23Added 13 answers

More generally,
 loga+nlogb=log(abn) n log(ab)=log(anbn)=log(ab)n

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