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