Question 958685
 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