Question

If {9}^{-1 + 2\,x} == \frac{{81}^{2+x}}{{3}^{x}} find x

Answer 1

$${9}^{-1 + 2\,x} = \frac{{81}^{2+x}}{{3}^{x}}$$

$$(3^2)^{-1 + 2\,x} = (3^4)^{2+x}\cdot3^{-x}$$

$$3^{-2+4x}=3^{8+4x-x}$$

$$-2+4x=8+3x$$

$$x=10$$