How do you write #log_4 (1/16) = -2# into its exponential form?

1 Answer
Feb 13, 2015

You use the definition of logarithm:
#log_ax=b => a^b=x#

So you can write your expression. taking the base #a# of the log (in this case is #4#) to the power of #b=-2# which is indeed equal to the argument of the log, i.e.:

#4^(-2)=1/4^2=1/16#

In which you use the fact that: #x^-a=1/x^a#

Alternatively you can take all as exponents of #4#:
#4^(log_4(1/16))=4^(-2)#
Which is again:
#1/16=1/16#