$3\log_{10}(x-15) = \left(\frac{1}{4}\right)^x$
I am completely lost on how to proceed. Could someone explain how to find any real solution to the above equation?
$3\log_{10}(x-15) = \left(\frac{1}{4}\right)^x$
I am completely lost on how to proceed. Could someone explain how to find any real solution to the above equation?
Hint: Consider left and right sides at $x=16$ and $x=17$. You won't find an explicit "closed-form" solution, but you can prove that it exists.
Put \begin{equation*} f(x) = 3\log_{10}(x - 15) - \left(\dfrac{1}{4}\right)^x. \end{equation*} We have $f$ is a increasing function on $(15, +\infty)$. Another way, $f(16)>0 $ and $f(17)>0$. Therefore the given equation has only solution belongs to $(16,17)$.
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, $1.7870412309 \cdot 10^{-10}$.