Moe logs

$\Large\log_a{b}=\log_b{a}$

$\color{#FFFFFF}{\text{I hate logarithms}}$

Let $\color{#EC7300}{y=\log_a{b}=\log_b{a}} \\ a=b^{\log_a{b}}\qquad b=a^{\log_b{a}} \\ a=\color{#3D99F6}{b^y}\qquad\quad\; b=\color{#3D99F6}{a}^y \\ b=(\color{#3D99F6}{b^y})^y \\ b=b^{y^2} \\ 1=\color{#EC7300}{y}^2 \\ \big(\log_a{b}\big)^2=1=\big(\log_b{a}\big)^2 \\ \sqrt{\big(\log_a{b}\big)^2=1=\big(\log_b{a}\big)^2} \\ \log_a{b}=\pm 1=\log_b{a}$

$\color{#FFFFFF}{\text{I hate logarithms}}$

Case 1: $\color{#EC7300}{y}=1 \\ \log_a{b}=1=\log_b{a} \\ \log_a{b}=1\qquad\log_b{a}=1 \\ b=a\qquad\qquad a=b$

$\quad$ Obviously a contradiction, since $a\neq b$

$\color{#FFFFFF}{\text{I hate logarithms}}$

Case 2: $\color{#EC7300}{y}=-1 \\ \log_a{b}=-1=\log_b{a} \\ \log_a{b}=-1\qquad\log_b{a}=-1 \\ b=\dfrac{1}{a} \qquad\qquad\;\, a=\dfrac{1}{b}$

$\color{#FFFFFF}{\text{I hate logarithms}}$

Which implies $\Large\color{#20A900}{ab=\boxed{1}}$

$\text {Let } x = log_ab ; a, b \in \mathbb {R}^+ ; a ≠ 0 ; b ≠ 0$

Then (due to the change of base theorem):

$log_ab = log_ba \iff x = \frac {1}{x} \iff x^2 = 1 \iff x = ±1$

Since a and b are distinct (a≠b, x≠1), therefore:

x = -1

$log_ab = -1$

$a = \frac {1}{b}$

$ab = \frac {1}{b} × b = \boxed {1}$