How do you evaluate #f(x)=log_16x# for x=4?

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Answer 1 · 2017-08-01T04:42:50

#log_16(4) = color(blue)(1/2#

We're asked to evaluate

#log_16(4)#

This can also be written as

#16^x = 4#

We know that #4# is the square root of #16#, so

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

(which is equivalent to #sqrt16 = 4#)

Therefore,

#log_16(4) = color(blue)(1/2#