Is it possible to solve this equation with logarithms and exponents? $$-\frac{1}{3}\log(4x-12)+6=\left(-\frac{1}{2}\right)^x $$
Out of all the logarithm laws I've learned (which is pretty limited), I have not found a way to solve for what x is yet. Can someone verify that this equation can be solved, and provide a few hints and pointers on the method/how to do so?

for $$x=2m$$ and $m\in \mathbb{Z}$ we get the equation
$$-\frac{1}{2}\log(8m-12)+6=\left(\frac{1}{4}\right)^m$$
solving that we get $$m=1.25\cdot 10^{17}$$
in the other case we set $$x=2n+1$$ and we have
$$-\frac{1}{2}\log(8n-8)+6=\left(-\frac{1}{2}\right)\left(\frac{1}{4}\right)^n$$
and we also have $$1.25\cdot 10^{17}$$ maybe both results are not exact.
