Mathematics, 15.01.2021 08:40, milkshakegrande101
Given the equation log2x + log2(x – 7) = 3, solve for x.
Answers: 2
![\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](/tpl/images/1038/4861/bd8af.png)
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Given the equation log2x + log2(x – 7) = 3, solve for x....
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