If $a, b,$ and $c$ are distinct positive real numbers different from $1$ such that $a^2 + b^2 = c^2$

Question

If $a, b,$ and $c$ are distinct positive real numbers different from $1$ such that $a^2 + b^2 = c^2$, find the value of $\frac{1}{\log_{c+b} a} + \frac{1}{\log_{c-b} a}$.

Accepted Answer

$2$ Explanation: First, apply the change-of-base reciprocal rule $\frac{1}{\log_x y} = \log_y x$ to rewrite the terms:
$\log_a (c+b) + \log_a (c-b)$.

Using the product property of logarithms:
$= \log_a [(c+b)(c-b)] = \log_a (c^2 - b^2)$.

We are given the Pythagorean identity relation $a^2 + b^2 = c^2 \implies c^2 - b^2 = a^2$.

Substitute $a^2$ into the expression:
$= \log_a (a^2) = 2 \log_a a = 2$.

Options

Explanation