Is it correct? $n^{(\log\,x)} = x^ {(\log\,n)} $?

Question

Is it correct? $$n^{(\log\,x)} = x^ {(\log\,n)} $$ Can you proof and describe that, for any base? Please explain completely. Thank you.

http://math.stackexchange.com/questions/794187/number-raised-to-log-expression — lab bhattacharjee, Aug 22 '14 at 16:05

score 7 · Answer 1 · answered Aug 22 '14 at 12:29

7

Hint: compute the logarithm of both sides. Since the logarithm function is one-to-one this will tell you if they are equal.

answered Aug 22 '14 at 12:29

Umberto P.

score 7 · Answer 2 · answered Aug 22 '14 at 12:35

7

$ n^{(\log x)} = (e^{\log n})^{(\log x)} = e^{(\log n)(\log x)} = e^{(\log x)(\log n)}= (e^{\log x})^{(\log n)} = x^{(\log n)} $

If you use $\log$ to a different base $b$, then use $b$ instead of $e$.

answered Aug 22 '14 at 12:35

lhf

2 Answers2