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?

• That looks like one equation, not a system. – GEdgar Dec 30 '14 at 19:21
• remember it is assumed $10^\text{variable} = 4x - 12$ – Irrational Person Dec 30 '14 at 19:22
• To what base are the logarithms taken? Is $x$ supposed to be an integer? [or how is the power of a negative number defined] – Mark Bennet Dec 30 '14 at 19:23
• What is $x$, real? How do you want to define $-1/2$ to irrational power? – GEdgar Dec 30 '14 at 19:24
• Well this is only Grade 12, so my best guess it only real and rational numbers. Both equations are separate, and I'm trying to find the point of intersection. The log is of base 10, the common logarithm. – L to the V Dec 30 '14 at 19:28

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.
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.