Question

log4 x^2 + logx^6 64 = 2

Answer 1

x≠1; x>0;

$\displaystyle log_{2^2} x^2 + log_{x^6}2^6 = 2;\\log_2x + log_x2 = 2;\\log_2x + \frac{1}{log_2x} = 2;\ *log_2x;\ log_2x\neq 0; x\neq 1;\\log^2_2x + 1 - 2log_2x=0;\ t=log_2x;$

t²-2t+1=0;

t=4-4=0;

t=(2)/2=1;

log₂x=1;

x=2;