How do you evaluate #log_4 (1/2)#?

Question

11537 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-07T20:00:30

#log_4(1/2) = -1/2#

Using the following properties:

we have

#log_4(1/2) = log_4(4^(-1/2))#

#= -1/2log_4(4)#

#= -1/2(1)#

#= -1/2#

Answer 2 · 2016-01-07T20:03:34

#log_4(1/2)=(-1/2)#

Based on the meaning of #log# and exponents:
If
#color(white)("XXX")log_4(1/2) = c#
then
#color(white)("XXX")4^c= 1/2#

We know that
#color(white)("XXX")4^c>=1# for #c>=0#
So
#color(white)("XXX")4^c=1/2 rarr c<0#

Also
#color(white)("XXX")4^((-k)) = 1/(4^k)#

Which leads us to asking for what value of #k# does
#color(white)("XXX")1/(4^k) = 1/2#?

Since
#color(white)("XXX")4^(1/2) = sqrt(4) = 2#

#1/2 = 1/(4^(1/2)) = 4^(-1/2)#
and
#color(white)("XXX")c=(-1/2)#