How do you evaluate #log_( 1/4)16#?

1 Answer
Jul 1, 2016

Answer:

#log_(1/4)16=color(blue)(-2)#

Explanation:

Remember that #log_b a = c# means #b^c=a#
and that #b^c=(1/b)^(-c)# (provided #c!=0#)

If #log_(1/4) 16 =c#
then
#color(white)("XXX")(1/4)^c=16#

#rArrcolor(white)("XXX")4^(-c)=16#

#rArrcolor(white)("XXX")-c=2#

#rArrcolor(white)("XXX")c=-2#

Therefore
#color(white)("XXX")log_(1/4) 16= -2#