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Working with other Bases

You can choose almost any number as the base for logarithms and exponentials. Since we have ten fingers we often think of 10x and the corresponding logarithm function $\log_{10} x$. In the old days, schoolchildren got lots of practice using tables of logarithms with base ten. If we had two fingers we might have used base two. Come to think of it, we do have two arms. Anyway, these days we only use one finger to peck at the keys of a calculator and the function $\log_{10} x$ is disappearing from the world.

Let's look at the derivatives of bx and $\log_b x$. In particular, let's start with 2x. Now you might be tempted to say

\begin{displaymath}\frac{d}{dx}(2^x) = x 2^{x-1} \text{\ \ \ \ BUT THAT IS WRONG} .\end{displaymath}

You have to learn to control your baser instincts. The power rule doesn't apply when the base  is a constant and the exponent is a variable. Instead, the correct answer is:

\begin{displaymath}\frac{d}{dx} (2^x) = 2^x \ln 2 .\end{displaymath}

A factor of $\ln 2$ multiplies the derivative. Why $\ln 2$, for goodness sake, you ask?

We will show you two different ways to obtain this answer. You can decide which you like better and stick to that one.


Method 1. Well, look. $2 = e^{\ln 2} $, so that $ 2^x = ({e^{
\ln 2 }})^x = e^{x(\ln 2) }=e^{\ln {(2^x)}}$ by the rules for logarithms. Applying the chain rule, we have

\begin{displaymath}\frac{d}{dx} 2^x = \frac {d}{dx} e^{\ln
{(2^x)}} = e^{\ln {(2...
...^x)}} \frac{d}{dx} x \ln 2 = e^{\ln {(2^x)}} \ln 2 = 2^x \ln 2.\end{displaymath}


Method 2. We want to find $\displaystyle \frac {dy}{dx}$ where y= 2x. Let's use logarithmic differentiation on that last equation.


y= 2x


\begin{displaymath}\ln y = ln {(2^x)}\end{displaymath}


\begin{displaymath}\ln y = x \ln 2\end{displaymath}


\begin{displaymath}\frac{1}{y} \frac{dy}{dx} = \ln 2\end{displaymath}


\begin{displaymath}\frac{dy}{dx} = y \ln 2 = 2^x ln 2\end{displaymath}

Just what we were expecting.


The same argument in either method gives the general case:


\begin{displaymath}\boxed{ \frac{d}{dx} (a^x) = a^x \ln a }\end{displaymath}


Now, what about differentiating logb x ?

We want to find $\displaystyle \frac {dy}{dx}$ where y = logb x.

But, y = logb x implies $\displaystyle b^y = x.$ Let's implicitly  differentiate this equation.


We get:


\begin{displaymath}b^y \ln b \frac{dy}{dx} = 1.\end{displaymath}


\begin{displaymath}\frac{dy}{dx} = \frac{1}{b^y \ln b} = \frac {1} {x \ln b }\end{displaymath}

We've just shown that:

\begin{displaymath}\boxed{ \frac{d}{dx} log_b x = \frac {1} {x \ln b }} .\end{displaymath}

An extra factor of $\ln b $ occurs for exponentials with base b. Notice that since $\ln e=1$, there is no extra factor exactly when the base is e. Makes you really appreciate e, doesn't it?



Next: Integrals and the Natural Up: Doing that Calc Thing Previous: Differentiating the Natural Log
Joel Hass
1999-05-26