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

Note that your equation is equivalent to $$ \frac{4^{9x - 4}}{3^{9x-4}} = 1 $$ or $$ \left(\frac{4}{3}\right)^{9x-4} = 1. $$

You want to solve
$$4^Q=3^Q$$ Which becomes

$$\left({4\over3}\right)^Q=1$$ Hence $$Q=0$$

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?

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 $$(9x-4)(\log 4-\log 3)=0$$

In general (for different exponents) you can change the base to $e$ to end up with a simpler equation: $$4^{9x-4}=3^{9x-4}$$ $$\implies e^{(9x-4)\ln4}=e^{(9x-4)\ln3}$$ $$\implies (9x-4)\ln4=(9x-4)\ln3$$ $$\implies x=\frac 49$$