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

Jun 23, 2017

$- 2$

Explanation:

We have: ${\log}_{\frac{1}{2}} \left(4\right)$

The change of base formula is ${\log}_{b} \left(a\right) = \frac{{\log}_{c} \left(a\right)}{{\log}_{c} \left(b\right)}$; 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:

$R i g h t a r r o w {\log}_{\frac{1}{2}} \left(4\right) = \frac{{\log}_{2} \left(4\right)}{{\log}_{2} \left(\frac{1}{2}\right)}$

$R i g h t a r r o w {\log}_{\frac{1}{2}} \left(4\right) = \frac{{\log}_{2} \left({2}^{2}\right)}{{\log}_{2} \left({2}^{- 1}\right)}$

Using the laws of logarithms:

$R i g h t a r r o w {\log}_{\frac{1}{2}} \left(4\right) = \frac{2 \cdot {\log}_{2} \left(2\right)}{- 1 \cdot {\log}_{2} \left(2\right)}$

$R i g h t a r r o w {\log}_{\frac{1}{2}} \left(4\right) = - \frac{2 \cdot 1}{1}$

$\therefore {\log}_{\frac{1}{2}} \left(4\right) = - 2$