4^x=32^(-x-2)*8^(x-3) find x

find x plssss

Solve for x over the real numbers:
4^x = 2^(-19-2 x)

Take the natural logarithm of both sides and use the identity log(a^b) = b log(a):
2 log(2) x = log(2) (-19-2 x)

Expand out terms of the right hand side:
2 log(2) x = -19 log(2)-2 log(2) x

Add 2 x log(2) to both sides:
4 log(2) x = -19 log(2)

Divide both sides by 4 log(2):
Answer: |x = -19/4

4^x=32^(-x-2)*8^(x-3) find x

$\begin{array}{rll}4^x&=&32^{-x-2}\times8^{x -3}\\4^x&=&\dfrac{8^x\times 8^{-3}}{32^x\times32^ 2}\\4^x&=&(\dfrac{1}{4})^x(\dfrac{1}{8^3\times 32^2}) \\\left(\dfrac{4}{\frac{1}{4}}\right)^x&=&\dfrac{1}{2^{19}}\\ 16^x&=&2^{-19}\\ 2^{4x}&=&2^{-19}\\ 4x &=& -19\\ x&=&-\dfrac{19}{4}\end{array}$

A method without logarithms. Similar to Omi67's answer.