Answer
$x=4$
Work Step by Step
$\Big(\dfrac{1}{8}\Big)^{x}=16^{1-x}$
Rewrite $8$ as $2^{3}$ and $16$ as $2^{4}$:
$\Big(\dfrac{1}{2^{3}}\Big)^{x}=(2^{4})^{1-x}$
Take $2^{3}$ to the numerator by changing the sign of its exponent:
$(2^{-3})^{x}=(2^{4})^{1-x}$
Multiply the exponents on both sides:
$2^{-3x}=2^{4-4x}$
If $2^{-3x}=2^{4-4x}$, then $-3x=4-4x$:
$-3x=4-4x$
Solve for $x$:
$-3x+4x=4$
$x=4$
