# How do you solve log(x+10)-log(x+4)=logx?

It is $x = 2$
log(x+10)-log(x+4)=logx=>log[(x+10)/(x+4)]=logx=> (x+10)/(x+4)=x=>x+10=x*(x+4)=> x^2+4x-x-10=0=> x^2+3x-10=0=> x_1=2 or x_2=-5
But hence $x > 0$ the only acceptable solution is $x = 2$