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: $a$ 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?
A: Although people have hinted at it, you must use the fact that any integer $n > 1$ has a unique prime factorization.  In order for to obtain $9^a = 15^b$, we must have $3^{2a} = 3^b5^b$, or $3^{2a-b} = 5^b$.  But by unique factorization, no positive power of $3$ can equal a positive power of $5$.  Thus, $\log_9 15$ must be irrational.
[Edited to account for the boundary case that $3^0 = 1 = 5^0$.]
A: Let's change the base of logarithm to $3$, i.e. $\log_9{15}=\dfrac{\log_3{15}}{\log_3{9}}=\dfrac{1+\log_3{5}}{2}$. 
Suppose the last expression is in $\mathbb{Q}$. Then $\log_3{5}\in \mathbb{Q}$. 
Then $\exists (a,b)\in \mathbb{Z}\times \mathbb{N}$ such that $3^a=5^b.$ But $5^b\equiv 0\quad\text{or}\quad5 \pmod{10}$ and $3^a\equiv 1,3,9,7\pmod{10}$. I think that here we get contradiction.
A: $$15^b = 9^a$$
After this, you can write it as
\begin{align*}
\frac{15^b}{9^a} & = 1\\
\frac{3^b5^b}{3^{2a}} & = 1\\
\frac{5^b}{3^{2a-b}} & = 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.
