Testwiki: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)