How do you solve #6^n=99#?

1 Answer
Sep 23, 2016

Answer:

I got: #n=2.56458#

Explanation:

I would try taking the log in base #6# of both sides:
#log_6(6^n)=log_6(99)#
we get (using the definition of log):
#n=log_6(99)#
the log is tricky but if we have a calculator we can evaluate it after changing its base, say, to get a natural log in base #e# indicated as #ln#. So we get:
#n=ln(99)/(ln(6))=2.56458#