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Answer 1 · 2017-02-17T22:20:57

We have that

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

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

#log[(2x+1)(x-2)]=1# [Use the additive law of logarithms]

#log[(2x+1)(x-2)]=log10# [We used the base 10 logarithm here]

#(2x+1)(x-2)=10#

#2 x^2 - 3 x - 12 = 0#

The last one (quadratic equation) has solutions

#x_1 = 3/4 - sqrt(105)/4# or #x_1≈-1.8117#

#x_2= 3/4 + sqrt(105)/4# or #x_2≈3.3117#

Because #(x-2)>=0# hence #x>=2# the acceptable solution is

#x_2= 3/4 + sqrt(105)/4#