Show that :

log_a m ÷ log_ab m = 1 + log_a b

Simplifying L.H.S. of the given equation, we get :

$\Rightarrow \text{log}$a \space m ÷ \text{log}{ab} \space m \\[1em] \Rightarrow \dfrac{\text{log m}}{\text{log a}} ÷ \dfrac{\text{log m}}{\text{log ab}}\\[1em] \Rightarrow \dfrac{\text{log m}}{\text{log a}} \times \dfrac{\text{log ab}}{\text{log m}}\\[1em] \Rightarrow \dfrac{\text{log ab}}{\text{log a}} \\[1em] \Rightarrow \text{log}{a} \space ab \\[1em] \Rightarrow \text{log}{a} \space a + \text{log}{a} \space b \\[1em] \Rightarrow 1 + \text{log}{a} \space b.

Hence, proved that log_a m ÷ log_ab m = 1 + log_a b.