Question

Evaluate log0.25 2.

Picture below:

Answer 1

A

Explanation:

using the property of logarithms

`log_(b)` x = n ⇒ x =
`b^(n)`

let
`log_(0.25)` 2 = n ( solve for n ) , then

2 =
`(0.25)^(n)` =
`((1)/(4)) ^(n)` =
`((1)/(2^2)) ^(n)` =
`(2^(-2)) ^(n)` =
`2^(-2n)`

so

`2^(-2n)` =
`2^(1)`

since bases on both sides are the same then equate exponents

- 2n = 1 ( divide both sides by - 2 )

n =
`(1)/(-2)` = -
`(1)/(2)`