How do you solve #4^(x-1)=3#?

1 Answer
May 5, 2015

You can try by taking the logarithm (of base #4#) of both sides:
#log_4(4^(x-1))=log_4(3)#
#log_4# and #4^()# cancel out, so you get:
#x-1=log_4(3)# and
#x=1+log_4(3)#

This can be a little tricky to solve unless you decide to change base and can use a calculator (or tables):
#log_4(3)=(ln3)/(ln4)~~0.8# (using natural logarithms)
So that:
#x=1.8#