Prove that $\log_a(b)=-\log_b(a)$

Question

Can you prove that: $$\log_a(b)=-\log_b(a)$$ I just thought that it should equal $$\frac{\log(b)}{\log(a)}.$$ but I don't think anything else.

Answer 1

$\log_a(b) = \frac{\log b}{\log a} = \frac{1}{\frac{\log a}{\log b}} = \frac{1}{\log_b(a)} ≠ -\log_b(a)$ in general.

Answer 2

let $$\log_a(b) = x \to b = a^x\to a= b^{1/x}\to \log_b(a) = \frac 1x=\frac1{\log_b(a)}$$ you can rewrite this as $$\log_a(b)\log_b(a) = 1.$$

in general we have the change of basis formula$$\log_a(b)\log_b(c) = \log_a(c).$$ the previous identity is the special case of $c = a.$