Question

log8(x^2-3x)=log8(5x-12)

Answer 1

$log_8(x^2-3x)=log_8(5x-12)\O.D.3.:\begin{cases}x^2-3x>0\5x-12>0end{cases}Rightarrow xin(3;;+infty)\\x^2-3x=5x-12\x^2-8x+12=0\D=64-4cdot12=16\x_{1,2}=frac{8pm4}2\x_1=6\x_2=1;-;He;nogx.;no;O.D.3.\\OTBET:;x=6$