Solve for $x$ in $2\log(x+11)=(\frac{1}{2})^x$ Solve for $x$.
$$2\log(x+11)=(1/2)^x$$
My attempt:
$$\log(x+11)=\dfrac{1}{(2^x)(2)}$$
$$10^{1/(2^x)(2)}= x+11$$
$$x=10^{1/(2^x)(2)}-11$$
I'm not sure what to do next, because i have one $x$ in the exponent while the other on the left side of the equation.

For $x > -11$ assume :
$k1(x) = 2log(x+11)$  which is a continuous, differential, monotonically 
increasing function with $k1'(x) > 0$
$k2(x) = (1/2)^x$ which is a continuous, differential, monotonically 
decreasing function with $k2'(x) < 0$
$K(x) = k1(x) - k2(x)$ which is a continuous, differential function.
$K'(x) = k1'(x) - k2'(x) > 0$ and so $K(x)$ is a monotonically increasing function.
We solve $K(x) = 0 \iff K(x) = K(-1)$.
K(x) is continuous and monotonic and so the obvious solution $x =-1$ is unique.
