Solve $3\log_{10}(x-15) = \left(\frac{1}{4}\right)^x$ $$3\log_{10}(x-15) = \left(\frac{1}{4}\right)^x$$
I am completely lost on how to proceed. Could someone explain how to find any real solution to the above equation?

Let $x=16+y$. After approximating $\log(1+y)$ with $y - \frac{y^2}{2}$, $(\frac{1}{4})^y = \exp(-\log(4) y)$ with $1 - \log(4) y$, $\frac{1}{1+\epsilon}$ with $1 - \epsilon$, get 
$$y \approx \frac{\log(10)}{3 \cdot 4^{16}}.$$
WIMS Function Calculator gives the exact solution, $1.7870412306 \cdot 10^{-10}$ compared to the approximation, $1.7870412309 \cdot 10^{-10}$.
