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