How do you write #3log_(4)16=2# in exponential form?

1 Answer
Sep 30, 2015

#4^2=16#

Explanation:

According to the definition of log:
#y=log_bx# is equivalent to #b^y=x#

So we have:
#3log_(4)(16)=2#
#log_(4)(16^3)=2#

Compare what we have to the definition of log:
#y=2#
#b=4#
#x=16^3#

Therefore:
#4^2=16^3#

However, #4^2# is definitely not equal to #16^3#. I think you probably meant "#log_4(16)=2#" instead, which would have resulted to the answer #color(blue)(4^2=16)#.