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How do you solve #log_3 x-log_3(x-1)=1#?

1 Answer

Aug 17, 2016

#x = 3/2#

Explanation:

#log_3 x-log_3(x-1)=1 = log_3 3# so
#log_3(x/(x-1))=log_3 3# so
#x/(x-1) = 3# or
#x =3(x-1)# Solving for #x# gives

#x = 3/2#. Verifying the feasibility

#log_3 (3/2) - log_3(1/2)=log_3 3#

So, #x = 3/2# is the solution.

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