I know that ${\displaystyle a^{\log _{b}(n)}=n^{\log _{b}(a)}}$(because I asked Wolfram Alpha), but I don't know why. I suspect it has something to do with ${\displaystyle a^{\log _{b}(n)}=a^{\ln(n)/\ln(b)}=a^{\ln(n){\frac {1}{\ln(b)}}}}$, but that is where I lose course. --Stephan Schulz (talk) 07:02, 11 June 2023 (UTC)[reply]

Take the log: With ${\displaystyle \ln x^{y}=y\ln x}$both sides end up as ${\displaystyle {\frac {\ln a\ln n}{\ln b}}}$. --Wrongfilter (talk) 07:16, 11 June 2023 (UTC)[reply]

Thanks a lot. I can follow that computation. It does not quite give me the the intuitive understanding I have hoped for (I probably need to do more work with logarithms), but I think I can convert it to a forward argument, which may give me more insight. --Stephan Schulz (talk) 07:44, 11 June 2023 (UTC)[reply]

The essence is the exponent rule ${\displaystyle c^{uv}=\left(c^{u}\right)^{v}=\left(c^{v}\right)^{u}.}$Substitute ${\displaystyle \log _{c}a}$and ${\displaystyle \log _{c}b}$for ${\displaystyle u}$and ${\displaystyle v}$in the last equation, and you get ${\displaystyle \left(c^{\log _{c}a}\right)^{\log _{c}b}=\left(c^{\log _{c}b}\right)^{\log _{c}a},}$which is the same as ${\displaystyle a^{\log _{c}b}=b^{\log _{c}a}.}$Walking this through with ${\displaystyle \log _{2}}$on a concrete example:

${\displaystyle 8^{\log _{2}32}=(2^{3})^{5}=(2^{5})^{3}=32^{\log _{2}8}.}$

 --Lambiam 09:42, 11 June 2023 (UTC)[reply]

To clarify, Lambiam's statement applied to the equivalence talked about above is this:

${\displaystyle a^{\log _{b}(n)}=b^{\log _{b}(a)\log _{b}(n)}=b^{\log _{b}(n)\log _{b}(a)}=n^{\log _{b}(a)}}$.

Another way of showing it, switching bases only once but also using more complicated log identities, is: ${\displaystyle a^{\log _{b}(n)}=a^{\frac {\log _{a}(n)}{\log _{a}(b)}}=n^{\frac {1}{\log _{a}(b)}}=n^{\log _{b}(a)}}$. GalacticShoe (talk) 13:51, 11 June 2023 (UTC)[reply]

The shortest version I can think of now uses ${\displaystyle x^{y}=e^{y\ln x}}$and includes some of the arguments above: ${\displaystyle a^{\log _{b}(n)}=e^{\left({\frac {\ln a\ln n}{\ln b}}\right)}=n^{\log _{b}a}}$. --Wrongfilter (talk) 15:52, 11 June 2023 (UTC)[reply]