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

Question

5903 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-25T00:38:21

It is $x=2$

It is

$log(x+10)-log(x+4)=logx=>log[(x+10)/(x+4)]=logx=> (x+10)/(x+4)=x=>x+10=x*(x+4)=> x^2+4x-x-10=0=> x^2+3x-10=0=> x_1=2 or x_2=-5$

But hence $x>0$ the only acceptable solution is $x=2$

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