How do you solve #log (x – 2) + log x = log 3#?

1 Answer
Daniel L.
Nov 28, 2015

This equation has one solution #x=3#

Explanation:

We start with: #log(x-2)+logx=log3#

Before calculating #x# we have to find the domain of the equation.
Since #log# is only defined for positive values we have to find out where #x-2>0#, so we get the domain: #D=(2;+oo)#

#logx*(x-2)=log3#

#x*(x-2)=3#

#x^2-2x=3#

#x^2-2x-3=0#

#Delta=4-4*1*(-3)=4+12=16#

#sqrt(Delta)=4#

#x_1=(2-4)/2=-1#

#x_2=(2+4)/2=3#

From those values only #x_2# is in the domain #D# so it is the only solution of the equation.

Answer: #x=2#

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