How do you evaluate #log_16 (1/2)#?

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Answer 1 · 2018-05-03T21:41:05

#-1/4#

#log_16(1/2) = x#

We know that #log_b(x) = y# equals to #b^y = x#. Therefore, we can rewrite it as:
#16^x = 1/2#

Make both sides have a base of #2#:
#2^(4x) = 2^-1#

Since both sides have the same base, it becomes:
#4x = -1#

Divide both sides by #color(blue)4#:
#(4x)/color(blue)4 = -1/color(blue)4#

Therefore,
#x = -1/4#

The answer is #-1/4#.

Hope this helps!