Answer:
(c) x = 2 and x = -8
Step-by-step explanation:
The rules of logarithms let you rewrite this as a quadratic equation. That equation will have two (2) potential solutions. We know from the domain of the log function that any negative value of x will be an extraneous solution.
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The rules of logarithms that apply are ...
[tex]\log(a)+\log(b)=\log(ab)\qquad\text{all logarithms to the same base}\\\\\log_b(a)=c\ \Longleftrightarrow\ b^c=a\\\\[/tex]
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take antilogs
We can rewrite the equation so that only one logarithm is involved. Then we can take antilogs.
[tex]\log_4(x)+\log_4(x+6)=2\\\\\log_4(x(x+6))=2\qquad\text{using the first rule}\\\\4^2=x(x+6)\qquad\text{using the second rule}[/tex]
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solve the quadratic
Adding (6/2)² = 9 to both sides will "complete the square."
16 +9 = x² +6x +9 . . . . . . . add 9
25 = (x +3)²
±√25 = x +3 = ±5 . . . . . take the square root(s)
x = -3 ±5 = {-8, +2}
The two potential solutions are x = 2 and x = -8.
