# Different definitions of logarithm (Part 8)

There are two apparently different definitions of a logarithm that appear in the secondary mathematics curriculum:

1. From Algebra II and Precalculus: If $a > 0$ and $a \ne 1$, then $f(x) = \log_a x$ is the inverse function of $g(x) = a^x$.
2. From Calculus: for $x > 0$, we define $\ln x = \displaystyle \int_1^x \frac{1}{t} dt$.

In the previous posts of this series, we established the connection between these two apparently different ways of defining a logarithm.

In this post, I describe some natural consequences of Definition 2 as it relates to calculus involving logarithmic and exponential functions. None of the results that follow are new to the senior math majors taking my capstone course. However, they almost unanimously tell me that they never learned why these rules worked when they were taking the calculus sequence several semesters ago… instead, they were given these rules to just memorize without worrying about where they came from. (I don’t doubt that some of them were taught the logical development that’s shown below, but perhaps they forgot about it since they were never held responsible for this logical development on homework or exams when they were freshmen or sophomores.)

A pedagogical note: because none of these theorems should be a surprise to students, I’ll write the left-hand sides and the equal sign on the board and then leave the right-hand side blank. After my students tell me what the punch line of the theorem should be, then I’ll write the complete statement of the theorem on the board before proceeding to the proof. If my students can’t remember the punch-line (which has happened for Theorem 2 below), I’ll just start the proof until we reach the conclusion and then fill in the blank right-hand side of the theorem’s statement later. Theorem 1. $\displaystyle \frac{d}{dx} (\ln x) = \displaystyle \frac{1}{x}$

Proof. This is a consequence of the Fundamental Theorem of Calculus. Loosely stated, the derivative of an integral is the original function. Theorem 2. $\displaystyle \frac{d}{dx} (\log_a x) = \displaystyle \frac{1}{x \ln a}$

Proof. This is a consequence of the change-of-base formula: $\log_a x = \displaystyle \frac{\log_e x}{\log_e a} = \displaystyle \frac{\ln x}{\ln a}$ $\frac{d}{dx} \left( \log_a x \right) = \displaystyle \frac{d}{dx} \left( \frac{\ln x}{\ln a} \right)$

Since $\ln a$ is just a constant, we conclude $\frac{d}{dx} \left( \log_a x \right) = \displaystyle \frac{1}{\ln a} \cdot \frac{d}{dx} ( \ln x ) = \displaystyle \frac{1}{\ln a} \cdot \frac{1}{x}$ Theorem 3. $\displaystyle \frac{d}{dx} \left( a^x \right) = a^x \ln a$

Proof. This proof uses a standard trick. Suppose that $y = a^x$

so that $\log_a y = x$

We need to compute the derivative $\displaystyle \frac{dy}{dx}$. To this end, let’s apply implicit differentiation to this last equation: $\displaystyle \frac{1}{y \ln a} \cdot \frac{dy}{dx} = 1$ $\displaystyle \frac{dy}{dx} = y \ln a$ $\displaystyle \frac{d}{dx} \left( a^x \right) = a^x \ln a$

Technical Note: The above proof assumes that the derivative $\displaystyle \frac{dy}{dx}$ exists in the first place, and so, technically, implicit differentiation is inappropriate (even if it does produce the correct answer). However, the existence proof is quite difficult, requiring an advanced version of the Mean Value Theorem. So, for pedagogical reasons, I have absolutely no qualms about showing the above proof to students in a class that does not require second-semester real analysis as a prerequisite.

Pedagogical Note: After presenting this proof to my students, students are usually emotionally stunned. They usually tell me that they never saw how the derivative of an inverse function could be computed back when they were in the calculus sequence… it was just another formula that they had to memorize. So (time permitting), to let this idea sink in a little further, I’ll go off on a tangent (pun intended)…

…to find the derivative of an inverse trigonometric function: $y = \sin^{-1} x$ $\sin y = x$ $\cos y \cdot \displaystyle \frac{dy}{dx} = 1$ $\displaystyle \frac{dy}{dx} = \frac{1}{\cos y}$ $\displaystyle \frac{dy}{dx} = \frac{1}{\sqrt{1 - \sin^2 y}}$ $\displaystyle \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$

I’ll then invite students to figure out the derivatives of $y = \cos^{-1} x$ and $y = \tan^{-1} x$ on their own after class. Theorem 4. $\displaystyle \frac{d}{dx} \left( e^x \right) = e^x$

Proof. Let $a = e$ in Theorem 3. Alternatively, repeat the above argument for the inverse functions $e^x$ and $\ln x$.

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