SOLUTION: what is the solution set to the equation log2(3x+1) + log2(x+7)=5?

Question 958685: what is the solution set to the equation log2(3x+1) + log2(x+7)=5?
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
log2(3x+1) + log2(x+7)=5
log2(3x+1)(x+7)=5
(3x+1)(x+7)= 2^5
3x^2+21x+x+7= 32
3x^2+22x+7= 32
3x^2+22x-25= 0
3x^2-3x+25x-25= 0
(3x^2-3x)+(25x-25)= 0
3x(x-1)+25(x-1)= 0
(x-1)(3x+25)= 0
x = {-25,1}
throw out the negative solution (extraneous) leaving
x = 1
RELATED QUESTIONS
what are the reasons for these steps that go in the boxes correctly to this problem... (answered by Theo)

Solve the equation: log2(x+7) +... (answered by stanbon,MathLover1)

Solve the equation... (answered by ewatrrr)