Question

помогите пжл срочно

Answer 1

$1)2^{x+8}= \frac{1}{32}\\\\2^{x+8}= 2^{-5}\\x+8=-5\\x=-13\\\\2)\frac{3}{3^{x+2}}=1\\\\3^{x+2}=3\\x+2=1\\x=-1\\\\3)0,5^{2x}* 4^{2x}=0,5* 2^{3x-1}\\\\(0,5*4)^{2x}=0,5* 2^{3x}* 2^{-1}\\\\2^{2x}= 2^{-2}* 2^{3x}\\\\2^{2x}= 2^{3x-2}\\3x-2=2x\\x=2$

$4)\sqrt[3]{\sqrt{2^{x}} }=4\\\\2^{\frac{x}{6}} = 2^{2}\\\\\frac{x}{6} =2\\x=12\\\\5)(\frac{3}{8}) ^{3x-8} =(\frac{8}{3}) ^{8x-3}\\\\(\frac{3}{8}) ^{3x-8}=( \frac{3}{8}) ^{3-8x} \\3x-8=3-8x\\3x+8x=3+8\\11x=11\\x=1$