Right angled triangle log

Question

If $a,b$ and $c$($c$ is the hypotenuse) are sides of a right triangle then prove $$(\log_{c+b}a)+(\log_{c-b}a)=2(\log_{c+b} a )\cdot(\log_{c-b}a)$$ The bases are different so can't quite figure out what to do

Answer 1

Since $c^2-b^2=a^2$, one has$$\begin{align}\log_{c+b}(a)+\log_{c-b}(a)&=\frac{\log_{a}(a)}{\log_{a}(c+b)}+\frac{\log_{a}(a)}{\log_{a}(c-b)}\\&=\frac{\log_{a}(c-b)+\log_a(c+b)}{\log_a(c+b)\times\log_{a}(c-b)}\\&=\frac{\log_a(c^2-b^2)}{\log_a(c+b)\times\log_{a}(c-b)}\\&=\frac{\log_a(a^2)}{\log_a(c+b)\times\log_{a}(c-b)}\\&=\frac{2}{\log_a(c+b)\times\log_{a}(c-b)}\\&=2\cdot\frac{\log_a(a)}{\log_{a}(c+b)}\cdot\frac{\log_a(a)}{\log_{a}(c-b)}\\&=2\log_{c+b}(a)\times\log_{c-b}(a)\end{align}$$

Answer 2

Recall that $a^2+b^2=c^2$ and $\log_b a = \frac{\ln a}{\ln b}$ (actually the base $e$ is arbitrary, I use it for clarity). You are then asked to prove that $$\begin{align*} \frac{\ln a}{\ln(c+b)} + \frac{\ln a}{\ln(c-b)} & = 2 \frac{\ln^2 a}{\ln(c+b) \ln(c-b)} \\ \Leftrightarrow\ln(c-b) + \ln(c+b) & = 2\ln a \\ \Leftrightarrow\ln((c-b)(c+b)) & = \ln a^2 \\ \Leftrightarrow\ln(c^2-b^2) & = \ln a^2 \\ \Leftrightarrow c^2 - b^2 & = a^2 \\ \Leftrightarrow a^2+b^2 & = c^2 & \Box \end{align*}$$