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

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Answer 1 · 2018-06-27T02:37:44

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

#e^(ln(2x+1))=e^(2-ln(x))#

#2x+1=e^2/x#

#2x^2+x=e^2#

#2x^2+x-e^2=0#

#x=(-1+-sqrt((1+8(e^2))))/4#

#2*(-1-sqrt((1+8(e^2))))/4+1<0#

Therefore #ln(2*(-1-sqrt((1+8(e^2))))/4+1)# is not defined and not a valid input for x.

This means #x=(-1+sqrt((1+8(e^2))))/4#