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:
$$\begin{align}
4^{9x-4} &= 3^{9x-4} \\
\log_4(4^{9x-4}) &= \log_4(3^{9x-4}) \\
\end{align}$$
Where do I go from there?
Thanks!
 A: In general (for different exponents) you can change the base to $e$ to end up with a simpler equation:
$$4^{9x-4}=3^{9x-4}$$
$$\implies e^{(9x-4)\ln4}=e^{(9x-4)\ln3}$$
$$\implies (9x-4)\ln4=(9x-4)\ln3$$
$$\implies x=\frac 49$$
A: Notice, first, that the power of both the numbers is the same.
This simplifies matters to a huge extent. You need not take any logarithms.
Your equality is: $$4^{9x-4}=3^{9x-4}$$
You are allowed to multiply/divide numbers with similar powers and take the power of the result. This is one of the Rules of Exponents: 
$$\left({x^{\alpha}\over y^{\alpha}}\right)=\left({x\over y}\right)^{\alpha}$$
Doing so, you get:
$$\left({4\over3}\right)^{9x-4}=1$$
Any number raised to power $0$ equals $1$. This means that $9x-4$ should be equal to $0$, or
$$ 9x-4=0$$
thus, $$x=\frac49$$
This is the solution to the given equality.
A: You want to solve
$$4^Q=3^Q$$
Which becomes
$$\left({4\over3}\right)^Q=1$$
Hence $$Q=0$$     
A: $$
\begin{array}{rrl}
&(9x-4) \log 4 &= (9x-4) \log 3\\
\implies &(9x-4) (\log 4 -\log 3) &= 0\\
\implies &9x -4  &= 0 \mbox{ (as $\log 4 \neq \log 3$)}\\
\implies &x &= \frac{4}{9}
\end{array}
$$
A: For any $A^n = B^n$ where $A \ne B$, $n = 0$
So:
$$9x - 4 = 0$$
$$x = 4/9$$
A: Magic property of logarithms:
$$\log_b(x^y) = y\log_b(x)$$
So here,
$$\log_4(4^{9x - 4}) = (9x-4)\log_4(4)$$
and
$$\log_4(3^{9x-4}) = (9x-4)\log_4(3)$$
so that
$$(9x-4)\log_4(4) = (9x-4)\log_4(3).$$
Now can you solve it?
A: So you thought @fasttouch was complex?
Adapted from Mathematica:
$$x =  \frac{4\log\frac{4}{3}- 2 \pi ni}{9 \log \frac{4}{3}},   n \in \Bbb Z$$
A: Note that your equation is equivalent to
$$
\frac{4^{9x - 4}}{3^{9x-4}} = 1
$$
or
$$
\left(\frac{4}{3}\right)^{9x-4} = 1.
$$
A: Note you can take logs to any base for this (provided it is the same both sides).
You get $$(9x-4)\log 4 = (9x-4)\log 3$$
which you can rewrite as $$(9x-4)(\log 4-\log 3)=0$$
A: 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 $4>3$ it follows that for  $Q \in \mathbb{N}^0$ and $Q>0$ it is true that $4^Q>3^Q$. Actually we can generalize that and we can say that for $Q>0$ and $Q\in\mathbb{R}$ it is still true.
Now for $Q<0$ you can write your equation in this way
$$
\frac{1}{4^{|Q|}} = \frac{1}{3^{|Q|}}
$$
and being $1/4<1/3$ 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?
A: This was actually my first train of thought and though not simplistic, it is another viewpoint.
$$
\begin{align*}
4^{9x-4} &= 3^{9x-4}\\
\log_4(4^{9x-4}) &= \frac{\log_3(3^{9x-4})}{\log_34}\\
9x-4 &= \frac{9x-4}{\log_34}\\
9x-4&=0\\
x &= \frac{4}{9}
\end{align*} 
$$
