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

1 Answer
Alan P.
Sep 20, 2016

#log_(1/4) (1/16)=color(green)(2)#

Explanation:

Remember:
#color(white)("XXX")log_color(blue)(b) color(magenta)(a) =color(red)(c) hArr color(blue)(b)^color(red)(c)=color(magenta)(a)#

Since
#color(white)("XXX")color(blue)(""(1/4))^color(red)(2) = color(magenta)(1/16)#
#rArr#
#color(white)("XXX")log_(color(blue)(1/4)) color(magenta)(1/16) = color(red)(2)#

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