Prove that $\log_9 15$ is irrational Im having trouble with the following proof... Ill post what I have completed so far..


  Prove that $\log_915$ is irrational.


Ill attempt by contradiction assuming $\log_915$ is rational.

So, 

$\log_915 = \frac ab$

$15 = 9^{\frac ab}$

$15^b = 9^a$
   (This is where I'm getting stuck)

Any hints/tips/advice would be great. Thanks
$a$ and $b$ are positive integers. $15^b$ and $9^a$ are positive integers. $5$ definitely does not divide $9^a$, so what must $b$ be?
$$15^b = 9^a$$

After this, you can write it as
\begin{align*}
\frac{15^b}{9^a} &amp; = 1\\
\frac{3^b5^b}{3^{2a}} &amp; = 1\\
\frac{5^b}{3^{2a-b}} &amp; = 1
\end{align*}

To fulfill this equation, numerator should be equal to denominator but no power of $3$ and $5$ will be same because power of $3$ will end with $1$,$3$,$7$,$9$ and that of $5$ will end with $5$ only, so their power can only be $0$.

$2b=0$ and $2a-b=0$

so $b$ equals zero but it is denominator in our first assumption of rationality, which contradicts our assumption.
Therefore, $\log_9 15$ is irrational.