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I was wondering if anyone has a proof for the logarithmic reciprocal property:

$\frac{1}{log_a(b)}=log_b(a)$

Thanks in advance, I sat down and tried to prove it myself but I couldn't do it and I haven't found any proof for this specific property online.

finnbratfisch
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The change of base formula says that $$\log_a(b) = {\log_c(b)\over \log_c(a)},$$ for any base $c > 0$.

ncmathsadist
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  • I'm having a hard time understanding how the reciprocal ties into this. – finnbratfisch Apr 28 '22 at 18:18
  • Set $c=a.$ Done. – Sean Roberson Apr 28 '22 at 18:19
  • I did and that just leaves me with $log_a(b)=\frac{log_a(b)}{log_a(a)}$ which simplifies to $log_a(b)=\frac{log_a(b)}{1}=log_a(b)$ – finnbratfisch Apr 28 '22 at 18:27
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    @finnbratfisch Set $c=b$. Note: $$\begin{align}b&=b\ a^{\log_a b}&=b\ \left(c^{\log_c a}\right)^{\log_a b}&=c^{\log_c b}\ c^{(\log_c a)(\log_a b)}&=c^{\log_c b}\ \therefore\quad (\log_c a)(\log_a b)&=\log_c b \end{align}$$ – r.e.s. Apr 28 '22 at 18:35
  • Yes, I see now. When $c=b$ this equation can be written as $(log_c(a))(log_a(b))=log_b(b)=1$. Dividing by $log_a(b)$ leaves you with $log_c(a)=\frac{1}{log_a(b)}$. Because $c=b$ this can be written as $log_c(a)=\frac{1}{log_a(c)}$. Thank you very much! – finnbratfisch Apr 28 '22 at 18:48