Change-of-Base Formula and Computing Logs with a Calculator
The Change-of-Base Formula
You have probably wondered at some point over the last few lessons “how can I plug something like \(log_3(2)\) into my calculator.” It is true that most calculators don’t have a generic “log” button that allows you to choose the base and argument (i.e. the thing inside parentheses), so we need a workaround for those of us with more primitive devices.
So basically the change-of-base formula allows us to take any arbitrary log of any base and turn it into a fraction of natural logs. This is good because there IS (very likely) a “\(ln\)” button on your calculator.
Using the change-of-base formula, note that
$$\begin{align}\log_2{3}=\frac{\ln(3)}{\ln(2)}\end{align}$$
Then, plugging each of \(\ln(3)\) and \(\ln(2)\) into your calculator, you get
$$\ln(3)=1.0986123\ \ \ \ and\ \ \ \ \ln(2)=0.693147$$
So we have
$$\begin{align}\log_2{3}=\frac{\ln(3)}{\ln(2)}=1.584963\end{align}$$
Note also that there is nothing special, really, about the use of natural logs in the change-of-base formula. In fact we can use any base we like! This has significance in many algebraic contexts where cancellation of exponential expression bases is called for. We won’t worry too much about that right now though.
Use the Change of Base formula to simplify or compute the following.
$$1.584962$$
$$\begin{align}\log_2(3)&=\frac{\ln(3)}{\ln(2)}\\&=\frac{1.098612}{0.6931471}\\&=1.584962\end{align}$$
$$2.153382$$
$$\begin{align} \log_5(32)&=\frac{\ln(32)}{\ln(5)}\\&=\frac{3.4657359}{1.6094379}\\&=2.153382\end{align}$$
$$-4.523562$$
$$\begin{align}\log_{\frac{1}{2}}(23)&=\frac{\ln(23)}{\ln(1/2)}\\&=\frac{3.135494}{-0.693147}\\&=-4.523562\end{align}$$