Question

$0,2 ^{x+1} = \sqrt{125}$

Answer 1

$(0,2)^{x+1}= \sqrt{125} \\\\ \left( \dfrac{2}{10}\right)^{x+1}= \sqrt{5^3}\\\\ \left( \dfrac{2\div2}{10\div2}\right)^{x+1}= \sqrt[2]{5^3}\\\\ \left( \dfrac{1}{5}\right)^{x+1}=5^{ \tfrac{3}{2} }\\\\ \left( \dfrac{1}{5^1}\right)^{x+1}=5^{ \tfrac{3}{2} }\\\\ (5^{-1})^{x+1}=5^{ \tfrac{3}{2} }\\ \not5^{-x-1}=\not5^{ \tfrac{3}{2}}\\\\ -x-1= \dfrac{3}{2}\\\\ 2\cdot(-x-1)=3\\ -2x-2=3\\ -2x=3+2\\ -2x=5\\\\ x=- \dfrac{5}{2}\\\\\\ S=\left\{- \dfrac{5}{2}\right\}$