# How do you evaluate log_(1/2) 5 using the change of base formula?

Sep 16, 2017

#### Answer:

I got: $- 2.32193$

#### Explanation:

Consider a log such as:

${\log}_{b} a$

we can change into the new base $e$ as:

${\log}_{b} a = \textcolor{red}{\frac{{\log}_{e} a}{{\log}_{e} b}}$

as you can see the new base gives a fraction between two new logs BUT we can easily use our calculator to evaluate ${\log}_{e}$ that is called the Natural Logarithm and is indicated as $\ln$. So we get:

${\log}_{b} a = \frac{{\log}_{e} a}{{\log}_{e} b} = \ln \frac{a}{\ln} \left(b\right) =$

in our case:

$\ln \frac{5}{\ln} \left(\frac{1}{2}\right) = \frac{1.60943}{- 0.69315} = - 2.32193$

[I forgot, you can test your result by applying the definition of log:
we got that:
${\log}_{\frac{1}{2}} \left(5\right) = - 2.32193$

so:
${\left(\frac{1}{2}\right)}^{- 2.32193} = 5$
that can be evaluated with the calculator]