<PRE><B>Question 254819</B>
Given, log2(x2 + 4x + 3) – log 2(x2 + x) = 4
       log2[(x^2+4x+3)/(x^2+x)] = 4
Taking anti-log, we get
           [(x^2+4x+3)/(x^2+x)] =2^4
             x^2+4x+3 = 16(x^2+x)
             x^2+4x+3 = 16x^2+16x
                    0 = 15x^2+12x-3 
Dividing by 3, we get
             5x^2+4x-1 = 0
             (5x-1)(x+1)=0
                      x = -1,1/5 
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