Prove that for any logarithim bases $a$ and $b$ $\log_ab=\dfrac 1{\log_ba}$ 
Prove that for any logarithim bases $a$ and $b$ $\log_ab=\dfrac 1{\log_ba}$

I am trying to prove by contradiction but I am stuck

Asssuming $$\log_ab\ne\dfrac 1{\log_ba}$$ Moving log to one side $$\log_ab\times\log_ba\ne1$$ 

How can I simplify this?
 A: Note: As this is absolutely trivial with the change of basis formula $\Big(\log_ab = \dfrac{\log_bb}{\log_ba} = \dfrac{1}{\log_ba}$. Done. $\Big)$ I'll assume that you haven't proved this formula yet.  So below I start from the definition of the logarithm in terms of exponentiation.
We know by definition that $$a^{\log_ab} = b \tag{1}$$ $$b^{\log_ba} = a\tag{2}$$ 
What we'd like here is to get rid of the exponent on one of the left hand sides.  If we could do that, we'd be able to plug it into the other equation.  So, remembering the exponent rule $(x^y)^z = x^{y\cdot z}$, let's exponentiate both sides of equation $(2)$ by $\dfrac{1}{\log_ba}$. Then we get $$(b^{\log_ba})^{1/\log_ba} = a^{1/\log_ba} \\ \text{but also } (b^{\log_ba})^{1/\log_ba} = b^{{\log_ba}\cdot(1/\log_ba)} = b^1 = b \\ \implies b = a^{1/\log_ba}$$  Now plug this into $(1)$ to get $$a^{\log_ab} = a^{1/\log_ba}$$
Because the function $a^x$ is injective/ one-to-one, we can then conclude that $\log_ab = \dfrac 1{\log_ba}$, as required.
A: $\log_ab=\dfrac {\log b}{\log{a}}$
$\log_ba=\dfrac {\log a}{\log{b}}$
A: Let $c = \log_a b$ Then 
$$a = (a^c)^{1/c} = b^{1/c}.$$
Therefore $\log_b a = 1/c$.
