Question

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

Answer 1

`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`