Question

Determine a solução de: 16^x-y = 64
9^x · 27^y = 1

Answer 1

$16^{x-y}=64\\ 9^x.27^y=1\\\\ 16^{x-y}=64\\ (4^2)^{x-y}=4^3\\ 4^{2x-2y}=4^3\\ 2x-2y=3\\\\ 9^x.27^y=1\\ (3^2)^x.(3^3)^y=1\\ 3^{2x}.3^{3y}=1\\ 2x+3y=1\\\\ 2x-2y=3\\ 2x+3y=1\\\\ 2x-2y=3 \ \ .(-1)\\ 2x+3y=1\\\\ -2x+2y=-3\\ 2x+3y=1\\\\ 5y=-2\\ y=\frac{-2}{5}\\\\ 2x+3y=1\\ 2x+3.\frac{-2}{5}=1\\ 10x-6=5\\ 10x=11\\ x=\frac{11}{10}$