Wikipedia:Reference desk/Archives/Mathematics/2023 June 11

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June 11

Logarithms and exponentiation

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

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

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)

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

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

 --Lambiam 09:42, 11 June 2023 (UTC)

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

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

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

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