Question 35219
logx + log(x+2)=3 --> the logs being at base2.
log(x(x+2))=3
{{{ x(x+2) = 2^3 }}}
{{{ x^2 + 2x = 8 }}}
{{{ x^2 + 2x - 8 = 0 }}}
(x+4)(x-2) = 0
so x+4=0 or x-2=0
--> x=-4, x=2


Now log(negative number) is not allowed, so although x=-4 comes out of the quadratic, it is only x=2 that is valid.



jon