Solve $4^{9x-4} = 3^{9x-4}$ I am having some trouble trying to solve 

$$4^{9x-4} = 3^{9x-4}$$

I tried to make each the same base but then I'm becoming confused as to what to do next.

These are the steps I took:

$$\begin{align}
4^{9x-4} &amp;= 3^{9x-4} \\
\log_4(4^{9x-4}) &amp;= \log_4(3^{9x-4}) \\
\end{align}$$

Where do I go from there?

Thanks!
Note that your equation is equivalent to
$$
\frac{4^{9x - 4}}{3^{9x-4}} = 1
$$
or
$$
\left(\frac{4}{3}\right)^{9x-4} = 1.
$$
Gerry said correctly. Can you solve it for $Q$? Now if $Q \in \mathbb{N}^0$ then the only solution is for $Q=0$. Since $4&gt;3$ it follows that for  $Q \in \mathbb{N}^0$ and $Q&gt;0$ it is true that $4^Q&gt;3^Q$. Actually we can generalize that and we can say that for $Q&gt;0$ and $Q\in\mathbb{R}$ it is still true.

Now for $Q&lt;0$ you can write your equation in this way
$$
\frac{1}{4^{|Q|}} = \frac{1}{3^{|Q|}}
$$
and being $1/4&lt;1/3$ you get the opposite of what we discussed before. That means that the only solution is $Q=0$ and that means $9x-4=0$ or $x=4/9$ as user1772257 has pointed out.

Does it help you?