How do you solve #ln(2x-5) - ln(4x)= 2#?

1 Answer
Jul 27, 2015

I found: #x=5/(2(1-2e^2)#. But it cannot be accepted.

Explanation:

I would use one of the rules of logs as:
#log_ax-log_ay=log_a(x/y)# to get:
#ln((2x-5)/(4x))=2# solving the log (with #ln=log_e#) you get:
#(2x-5)/(4x)=e^2#
rearranging:
#2x-5=4xe^2#
#2x-4xe^2=5#
#2x(1-2e^2)=5#
so that:
#x=5/(2(1-2e^2))=-0.363#
BUT it is negative!
If you substitute back the arguments of the #ln# become negative!!!
We cannot accept them; so NO (real) SOLUTIONS!