How do you solve #log x + log (x - 3) = 1#?

1 Answer
SagarStudy
Jan 6, 2016

#x=5#

Explanation:

#logx+log(x-3)=1#
We know that: #loga+logb=log(a*b)#
#implies Log(x(x-3))=1#
#implies log(x^2-3x)=1#
#implies x^2-3x=10#
#implies x^2-3x-10=0#
#implies (x-5)(x+2)=0#
#implies x=5,-2#

Verification:-
Put #x=5#
#L.H.S=Logx+Log(x-3)=Log5+log(5-3)=log5+log2=log(5*2)=log10=1=R.H.S#
Verified.
Put #x=-2#
#L.H.S=Logx+Log(x-3)=Log(-2)+log(-2-3)=log(-2)+log(-5)#
Here we have to find the log of a negative number which is undefined.
Therefore not verified.

Therefore, only #x=5# is true.

Related questions
See all questions in Logarithmic Models
Impact of this question
1293 views around the world
You can reuse this answer
Creative Commons License