Question

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

Answer 1

`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)`