How do you solve #log(2x+1)+log(x-1) = log (4x-4)#?

1 Answer
Aug 1, 2015

#color(red)(x=3/2)#

Explanation:

#log(2x+1) + log(x-1) = log (4x-4)#

Recall that #log x + log y = log(xy)#, so

#log(2x+1)(x-1) = log(4x-4)#

Convert the logarithmic equation to an exponential equation.

#10^(log(2x+1)(x-1)) = 10^( log (4x-4))#

Remember that #10^logx =x#, so

#(2x+1)(x-1) = 4x-4#

Factor the right hand side.

#(2x+1)(x-1) = 4(x-1)#

Divide both sides by #x-1#, where #x≠1#.

#2x+1=4#

#2x=3#

#x=3/2#

Check:

#log(2x+1) + log(x-1) = log (4x-4)#

If #x=3/2#

#log(2(3/2)+1) + log(3/2-1) = log (4(3/2)-4)#

#log(3+1)+log(1/2) = log (6-4)#

#log4–log1-log2 = log 2#

#log(4/2) -0 = log 2#

#log2 = log2#

#x=3/2# is a solution.