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

1 Answer
Trevor Ryan.
Jan 16, 2016

From laws of logs we get

#ln[(x+1)/(x-2)]=lnx^2#

#therefore (x+1)/(x-2)=x^2#

#therefore x^3-2x^2-x-1=0#

This is now a 3rd degree polynomial so you will have to try using the factor and remainder theorem to find the roots, otherwise you will have to use a numerical technique such as Newton's method of root-finding.

I leave the details as an exercise. Please ask if you are still stuck.

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