Question

Given the equation log2x + log2(x – 7) = 3, solve for x.

Answer 1

x∈{-1, 8}

Explanation:

`\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=(-(-7)\pm√((-7)^2-4\cdot1\cdot(-8)))/(2\cdot1)= (7\pm√(49+32))/(2) =(7\pm√(81))/(2)\\\\x=\frac{7+9}2=8\qquad\vee\qquad x=\frac{7-9}2=-1`