Question

Resolva a equação logarítmica:
a) log4 (2-x)² - log4 (17-x) = 1

Answer 1

$log_4(2-x)^2-log_4(17-x)=1\\ \\ log_4(\frac{(2-x)^2}{17-x})=1\\ \\ \frac{(2-x)^2}{17-x}=4\\ \\ (2-x)^2=4(17-x)\\ \\ 4-4x+x^2=68-4x\\ \\ x^2=64\\ \\ x=\pm \sqrt{64}\\ \\ \boxed{x=\pm 8}$

Answer 2

$log_4(2-x)^2-log_4(17-x)=1 \\ \\C.E.: \\ \\ (2-x)^2>0 \\ x^2-4x+4>0 \\ x \neq 2 \\ \\17-x>0 \\ x<17 \\ \\\\\\log_4(2-x)^2-log_4(17-x)=1\\\\ log_4[\frac{(2-x)^2}{(17-x)}]=1 \\ \\4^1= \frac{(2-x)^2}{(17-x)} \\ \\ 4= \frac{x^2-4x+4}{17-x}\\\\4(17-x)=x^2-4x+4 \\ \\ 68-4x=x^2-4x+4 \\ \\ x^2-4x+4+4x-68=0 \\ \\ x^2-64=0 \\ \\ x^2=64 \\ \\ x= \sqrt{64} \\ \\ x=8 \\ ou \\ x=-8$

S={8, -8}