Question 1000774
The difference of logs is the log of a quotient, so that from
log(3+x)-log(x-4)=log2  we get
log[(3+x)/(x-4)]=log 2
Now exponentiate ten-to-the and get
(3+x)(x-4)= 2
x^2-x-12=2
x^2-x-14=0
Now apply the quadratic formula, since this doesn't factor...
x = [1 +- sqrt(57)]/2