Question

log2(х в квадрате-4х+4)=4

Answer 1

$log_2(x^2-4x+4)=4\\ x^2-4x+4\ \textgreater \ 0\\ (x-2)^2\ \textgreater \ 0\\ x \in R \backslash \{x=2\}\\ log_2(x^2-4x+4)=4*1\\ log_2(x^2-4x+4)=4*log_22\\ log_2(x^2-4x+4)=log_22^4\\ x^2-4x+4=16\\ x^2-4x-12=0\\ D=16-4*1*(-12)=64\\ x_1= \frac{4+8}{2}=6\\ x_2= \frac{4-8}{2}=-2$