SOLUTION: How do you solve this equation? 4^(log2(3+x)) = 2x^2+8x+10 [the 2 is a base]

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Question 852985: How do you solve this equation?
4^(log2(3+x)) = 2x^2+8x+10 [the 2 is a base]

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi,
4^(log2(3+x)) = 2x^2+8x+10
(2^2)^(log2(3+x)) = 2x^2+8x+10
(2)^(2log2(3+x)) =2x^2+8x+10
(2)^(log2(3+x)^2) = 2x^2+8x+10
(3+x)^2) = 2x^2+8x+10
9 + 6x + x^2 = 2x^2+8x+10
x^2 + 2x + 1 = 0 Factor
(x+1)(x+1) = 0 , x = -1
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