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

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: How do you solve this equation? 4^(log2(3+x)) = 2x^2+8x+10 [the 2 is a base]       Log On


   

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) About Me  (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     +2%5E%28log%282%2CV%29%29+=+V

        (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