An algebra problem by Ossama Ismail

$\displaystyle 100 \times 2017^{-x} = \frac {100}{2017^x}$

$\displaystyle 2017^{3x} = 64$

$\displaystyle log\, _{2017}\, 64 = 3x$

$\displaystyle log\, _{2017}\, 2^6 = 3x$

$\displaystyle 6\,log\, _{2017}\, 2 = 3x$

$\displaystyle 2\,log\, _{2017}\, 2 = x$

$\displaystyle log\, _{2017}\, 4 = x$

$\displaystyle 2017^x = 4$

$\displaystyle \frac {100}{4} = 25$

$2017^{3x}=64$

My goal is to isolate exponent $x$ . If we represent the right-hand side of the equation as a cube, it would make separating $x$ possible by cancelling the $3$ s with the exponents on both the sides.

$2017^{3x}=4^3$
$\sqrt[3]{2017^{3x}}=\sqrt[3]{4^3}$
$2017^x=4^1$

So, $2017^{-x}$ would be $2017^{-1\cdot x}$

$=(2017^x)^{-1}$

$=(4)^{-1}$

$=\frac{1}{4}$

Therefore,

$100 \times \frac{1}{4}$ $=\boxed{25}$