How do you evaluate #log_(1/2) 4# using the change of base formula?

1 Answer
Jun 23, 2017

Answer:

#- 2#

Explanation:

We have: #log_(frac(1)(2))(4)#

The change of base formula is #log_(b)(a) = frac(log_(c)(a))(log_(c)(b))#; where #c# is the desired base.

In our case, the base is #frac(1)(2)# and the argument is #4#, both of which can be expressed in terms of #2#.

So let's set #2# as the new base:

#Rightarrow log_(frac(1)(2))(4) = frac(log_(2)(4))(log_(2)(frac(1)(2)))#

#Rightarrow log_(frac(1)(2))(4) = frac(log_(2)(2^(2)))(log_(2)(2^(- 1)))#

Using the laws of logarithms:

#Rightarrow log_(frac(1)(2))(4) = frac(2 cdot log_(2)(2))(- 1 cdot log_(2)(2))#

#Rightarrow log_(frac(1)(2))(4) = - frac(2 cdot 1)(1)#

#therefore log_(frac(1)(2))(4) = - 2#