Calculus with Other Bases Revision Notes for HSC SSCE Mathematics Advanced

Calculus with Other Bases

Introduction

When working with exponential functions in calculus, we typically use base $e$ . However, sometimes you'll encounter functions with different bases, such as $y = 2^x$ or $y = 10^x$ . The change-of-base formula allows us to differentiate and integrate exponential and logarithmic functions of any base without converting to base $e$ first.

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Understanding how to work with different bases is essential because many real-world applications use bases other than $e$ . For example, population growth models often use base 2, and pH calculations use base 10.

In this note, we develop three standard formulas that make working with other bases straightforward. Throughout, the base $a$ must be positive and not equal to 1.

Logarithmic functions to other bases

The change-of-base formula

Any logarithmic function can be expressed in terms of $\log_e x$ using the change-of-base formula. This is important because it shows that every logarithmic function is just a constant multiple of the natural logarithm.

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Change-of-base formula:

$\log_a x = \frac{\log_e x}{\log_e a}$

This can also be written as:

$\log_a x = \frac{1}{\log_e a} \times \log_e x$

This relationship tells us that $\log_a x$ is simply $\log_e x$ multiplied by the constant $\frac{1}{\log_e a}$ . This means we can treat $\frac{1}{\log_e a}$ as a constant when differentiating, making the process straightforward.

lightbulbExample

Worked Example: Using the change-of-base formula

Question:

a) Express $y = \log_5 x$ in terms of $\log_e x$

b) Use the calculator's ln function to approximate these values to four decimal places:

$\log_5 30$
$\log_5 2$
$\log_5 0.07$

c) Check the results using the power function

Solution:

Part a:

Using the change-of-base formula:

$\log_5 x = \frac{\log_e x}{\log_e 5}$

Part b:

For $\log_5 30$ :

$\log_5 30 = \frac{\log_e 30}{\log_e 5} \approx 2.1133$

For $\log_5 2$ :

$\log_5 2 = \frac{\log_e 2}{\log_e 5} \approx 0.4307$

For $\log_5 0.07$ :

$\log_5 0.07 = \frac{\log_e 0.07}{\log_e 5} \approx -1.6523$

Part c:

We can verify these results by checking that $5$ raised to each power gives the original number:

$5^{2.1133} \approx 30$

$5^{0.4307} \approx 2$

$5^{-1.6523} \approx 0.07$

Differentiating logarithmic functions with other bases

Understanding the differentiation process

Once we've expressed a logarithmic function as a multiple of $\log_e x$ , we can differentiate it using the standard rules we already know. Because $\frac{1}{\log_e a}$ is a constant, we can apply the constant multiple rule.

Two approaches to differentiation

There are two equally valid ways to differentiate logarithmic functions with bases other than $e$ :

Method 1: Use the change-of-base formula to convert to base $e$ , then differentiate

Method 2: Use the standard form directly: $\frac{d}{dx}\log_a x = \frac{1}{x \log_e a}$

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Both methods give the same result, but Method 2 is often faster once you're familiar with the formula. Method 1 is useful for understanding where the formula comes from.

Deriving the standard form

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Worked Example: Deriving the differentiation formula

Question: Use the change-of-base formula to differentiate $y = \log_a x$ .

Solution:

Starting with $y = \log_a x$

Using the change-of-base formula:

$y = \frac{\log_e x}{\log_e a}$

Since $\log_e a$ is a constant:

$\frac{dy}{dx} = \frac{1}{x} \times \frac{1}{\log_e a}$

$= \frac{1}{x \log_e a}$

This gives us the general formula for differentiating logarithms with any base.

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Standard form for differentiating logarithms:

$\frac{d}{dx}\log_a x = \frac{1}{x \log_e a}$

Notice that when $a = e$ , we have $\log_e e = 1$ , so this reduces to $\frac{d}{dx}\log_e x = \frac{1}{x}$ , which is the familiar formula we already know.

Worked examples of differentiation

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Worked Example: Differentiating with different bases

Question: Differentiate:

a) $\log_{10} x$

b) $\log_{1.05} x$

Solution:

Part a:

$\frac{d}{dx}\log_{10} x = \frac{1}{x \log_e 10}$

Part b:

$\frac{d}{dx}\log_{1.05} x = \frac{1}{x \log_e 1.05}$

Characterisation of the logarithmic function

The gradient at the x-intercept

An interesting property distinguishes the natural logarithm from all other logarithmic functions: it's the only one whose gradient at the x-intercept equals exactly 1.

This special property helps explain why base $e$ is so fundamental to calculus. While other bases work perfectly well mathematically, base $e$ gives the simplest and most elegant formulas.

Proving the special property of base e

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Worked Example: The special gradient property

Question:

a) Show that the tangent to $y = \log_a x$ at the x-intercept has gradient $\frac{1}{\log_e a}$

b) Show that $y = \log_e x$ is the only logarithmic function whose gradient at the x-intercept is exactly $1$

Solution:

Part a:

Starting with $y = \log_a x$

When $y = 0$ :

$\log_a x = 0$

$x = 1$

So the x-intercept is at $(1, 0)$ .

Differentiating:

$y' = \frac{1}{x \log_e a}$

When $x = 1$ :

$y' = \frac{1}{\log_e a}$

as required.

Part b:

The gradient at the x-intercept is $1$ if and only if:

$\log_e a = 1$

$a = e^1$

$= e$

Therefore, the gradient equals $1$ only when the base $a$ equals $e$ .

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Key characterisation of the natural logarithm:

The function $y = \log_e x$ is the only logarithmic function whose gradient at the x-intercept is exactly $1$ . This special property is one reason why base $e$ is fundamental to calculus.

Exponential functions with other bases

Converting to base e

Before we can apply calculus to an exponential function $y = a^x$ with base $a$ different from $e$ , we must convert it to base $e$ . We use the important identity:

$e^{\log_e a} = a$

This identity expresses the fact that $e^x$ and $\log_e x$ are inverse functions. Just as $\log_e(e^x) = x$ , we also have $e^{\log_e a} = a$ for any positive number $a$ .

The conversion process

We can write $a^x$ in terms of base $e$ as follows:

$a^x = (e^{\log_e a})^x$

Using the index law $(e^k)^x = e^{kx}$ :

$= e^{x \log_e a}$

Now $a^x$ has been expressed in the form $e^{kx}$ where $k = \log_e a$ is a constant. This form is ready for differentiation and integration using our standard rules for exponential functions.

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Key formulas for converting to base e:

Every positive real number can be written as a power of $e$ :

$a = e^{\log_e a}$

Every exponential function can be written with base $e$ :

$a^x = e^{x \log_e a}$

These conversions are essential because all our differentiation and integration rules for exponential functions are based on $e^x$ .

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Worked Example: Converting to powers of e

Question: Express these numbers and functions as powers of $e$ :

a) $2$

b) $2^x$

c) $5^{-x}$

Solution:

Part a:

$2 = e^{\log_e 2}$

Part b:

$2^x = (e^{\log_e 2})^x = e^{x \log_e 2}$

Part c:

$5^{-x} = (e^{\log_e 5})^{-x} = e^{-x \log_e 5}$

Differentiating and integrating exponential functions with other bases

Deriving the standard forms

Once we've written the function as a power of $e$ , we can differentiate and integrate using our standard rules. The key is recognizing that $\log_e a$ is just a constant multiplier.

Starting with $a^x = e^{\log_e a \cdot x} = e^{x \log_e a}$

For differentiation:

$\frac{d}{dx}a^x = \frac{d}{dx}e^{x \log_e a}$

$= e^{x \log_e a} \times \log_e a$

$= a^x \log_e a$

For integration:

$\int a^x \, dx = \int e^{x \log_e a} \, dx$

$= \frac{e^{x \log_e a}}{\log_e a}$

$= \frac{a^x}{\log_e a} + C$

chatImportant

Standard forms for exponential functions with base a:

Two approaches for working with base a:

Convert all powers to base $e$ before differentiating or integrating
Use the standard forms directly:

$\frac{d}{dx}a^x = a^x \log_e a$

$\int a^x \, dx = \frac{a^x}{\log_e a} + C$

Key observation: Both formulas involve the constant $\log_e a$ . When differentiating, you multiply by $\log_e a$ . When integrating, you divide by $\log_e a$ .

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Why base e gives the simplest formulas:

Notice that both formulas involve the constant $\log_e a$ . This constant equals $1$ when $a = e$ , which is why the formulas are simplest with base $e$ :

$\frac{d}{dx}e^x = e^x \times 1 = e^x$
$\int e^x \, dx = \frac{e^x}{1} + C = e^x + C$

This confirms that $e$ is the most natural base for calculus with exponential functions.

lightbulbExample

Worked Example: Differentiating and finding gradient

Question: Differentiate $y = 2^x$ . Hence find the gradient of $y = 2^x$ at the y-intercept, correct to three significant figures.

Solution:

Starting with $y = 2^x$

Using the standard form:

$y' = 2^x \log_e 2$

When $x = 0$ (the y-intercept):

$y' = 2^0 \times \log_e 2$

$= \log_e 2$

$\approx 0.693$

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Worked Example: Area between curves

Question:

a) Show that the line $y = x + 1$ meets the curve $y = 2^x$ at $A(0, 1)$ and $B(1, 2)$

b) Sketch the two curves and shade the region between them

c) Find the area of this shaded region, correct to four significant figures

Solution:

Part a:

For point $A(0, 1)$ :

When $x = 0$ : $y = 0 + 1 = 1$ and $y = 2^0 = 1$ ✓

For point $B(1, 2)$ :

When $x = 1$ : $y = 1 + 1 = 2$ and $y = 2^1 = 2$ ✓

Part b:

The graph shows the line $y = x + 1$ intersecting the exponential curve $y = 2^x$ at the two points. The shaded region lies between them from $x = 0$ to $x = 1$ .

Part c:

The area is found by integrating the difference between the upper and lower curves:

$\text{Area} = \int_0^1 (\text{upper curve} - \text{lower curve}) \, dx$

$= \int_0^1 (x + 1 - 2^x) \, dx$

$= \left[\frac{1}{2}x^2 + x - \frac{2^x}{\log_e 2}\right]_0^1$

Evaluating at the limits:

At $x = 1$ :

$= \frac{1}{2} + 1 - \frac{2}{\log_e 2}$

At $x = 0$ :

$= 0 + 0 - \frac{1}{\log_e 2}$

Therefore:

$\text{Area} = \left(\frac{1}{2} + 1 - \frac{2}{\log_e 2}\right) - \left(0 + 0 - \frac{1}{\log_e 2}\right)$

$= 1\frac{1}{2} - \frac{1}{\log_e 2}$

$\approx 0.05730 \text{ square units}$

bookmarkSummary

Key Points to Remember:

Change-of-base formula: Any logarithmic function can be written as $\log_a x = \frac{\log_e x}{\log_e a}$ , making it a constant multiple of $\log_e x$
Differentiating logarithms: Use $\frac{d}{dx}\log_a x = \frac{1}{x \log_e a}$ , or convert to base $e$ first
Converting exponentials: Any exponential function can be written as $a^x = e^{x \log_e a}$ before applying calculus
Differentiating exponentials: Use $\frac{d}{dx}a^x = a^x \log_e a$ — remember to multiply by $\log_e a$
Integrating exponentials: Use $\int a^x \, dx = \frac{a^x}{\log_e a} + C$ — remember to divide by $\log_e a$
Special property of base e: The function $y = \log_e x$ is the only logarithmic function with gradient exactly $1$ at its x-intercept, which is why base $e$ is fundamental to calculus
Memory aid: "To differentiate $a^x$ , multiply by $\log_e a$ ; to integrate $a^x$ , divide by $\log_e a$ "

Calculus with Other Bases (HSC SSCE Mathematics Advanced): Revision Notes