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

1 Answer

Konstantinos Michailidis

Feb 25, 2016

It is #x=2#

Explanation:

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#

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