Prove that for all constants $a, b>1$, $\log_a{n}$ is $\Theta(\log_b{n})$

Question

So far I have established that I should be showing the following relationship:

$k_{1}$ $\leq\ \frac{\log_a{n}}{\log_b{n}} \leq\ k_{2}$ where $k_{1}, k_{2}$ $>$ 0

but I'm not sure how to go about this.

Any ideas?

Answer 1

According to the change of base formula, we have

$$\frac{\log_a(n)}{\log_b(n)}=\log_a(b)$$

And then notice that

$$0<\log_a(b)<b$$