How do you solve #ln(-3+3n)=ln(n^2-n)#?

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Answer 1 · 2016-08-10T08:31:26

#n = 3#

#log_e(-3+3n)-log_e(n^2-n) = log(n-1)+log_e 3-log_e(n-1)-log_e n = 0#

or

#log_e 3-log_e n = log_e 1 #

Finally

#log_e(3/n) = log_e 1#

or

#n/3 = 1->n = 3# which is a feasible solution