Answer:
x∈{-1, 8}
Step-by-step explanation:
[tex]\log_2x +\log_2(x-7) = 3\\\\\log_2[x\cdot(x-7)]= 3\log_22\\\\\log_2(x^2-7x)= \log_22^3\\\\x^2-7x=2^3\\\\x^2-7x-8=0\quad\implies \quad a=1\,,\ \ b=-7\,,\ \ c=-8 \\\\x=\dfrac{-(-7)\pm\sqrt{(-7)^2-4\cdot1\cdot(-8)}}{2\cdot1}= \dfrac{7\pm\sqrt{49+32}}{2} =\dfrac{7\pm\sqrt{81}}{2}\\\\x=\dfrac{7+9}2=8\qquad\vee\qquad x=\dfrac{7-9}2=-1[/tex]
