$$-\frac{1}{3}\log(4x-12)+6=\left(-\frac{1}{2}\right)^x $$
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?
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The whole idea is to notice that, due to the very slow-growing nature of the logarithmic function, combined with the highly-accelerated decrease of an exponential function with sub-unitary basis, the equation does not possess any small-valued solutions, thus becoming ultimately equivalent to solving $\log(4x-12)\approx6\cdot3=18\iff x\approx3+\dfrac{a^{18}}4$ , where a represents the unspecified basis of your logarithm, presumably either $2$, or $10$, or e.
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.