Chapter 4: Exponential and Logarithmic Functions

In the previous chapters, we focused on polynomial functions like $x^2$ and $x^3$. In Chapter 4, we explore functions where the variable $x$ is the exponent. These functions are essential for modeling population growth, radioactive decay, and compound interest.

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4.1 Exponential Functions and the Number $e$

An exponential function has the form $f(x) = a^x$. However, in Mathematical Methods, we focus primarily on the natural exponential base, $e$.

• The constant $e$: Approximately $2.71828$. It is a unique number because the gradient of the function $f(x) = e^x$ is exactly equal to the value of the function itself at any point.
• Asymptotes: The graph of $y = e^x$ never touches the $x$-axis; it has a horizontal asymptote at $y = 0$.

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4.2 Logarithmic Functions

The logarithm is the inverse of an exponential. If $y = e^x$, then $x = \ln(y)$.

• $\ln(x)$ (Natural Log): This is the logarithm to the base $e$.
• Domain and Range: You cannot take the log of a negative number or zero. Therefore, for $f(x) = \ln(x)$, the domain is $x > 0$.

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4.3 Logarithm Laws

To solve equations involving $e$ and $\ln$, you must master the log laws. These are the same regardless of the base, but we use them most often with the natural log:

1. Product Law: $\ln(ab) = \ln(a) + \ln(b)$
2. Quotient Law: $\ln(\frac{a}{b}) = \ln(a) – \ln(b)$
3. Power Law: $\ln(a^n) = n \ln(a)$
4. Inverse Properties: $\ln(e^x) = x$ and $e^{\ln(x)} = x$

Worked Example 1: Solving for $x$

Solve $5e^{2x} = 20$.

1. Divide by 5: $e^{2x} = 4$.
2. Take the natural log of both sides: $\ln(e^{2x}) = \ln(4)$.
3. Use the inverse property: $2x = \ln(4)$.
4. Solve for $x$: $x = \frac{\ln(4)}{2} \approx 0.693$.

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4.4 Derivatives of $e^x$ and $\ln(x)$

Calculus becomes very elegant when dealing with base $e$.

• The Derivative of $e^x$: $\frac{d}{dx}(e^{kx}) = ke^{kx}$ (Essentially, the function stays the same, but you multiply by the derivative of the exponent).
• The Derivative of $\ln(x)$: $\frac{d}{dx}(\ln(x)) = \frac{1}{x}$

Worked Example 2: Differentiating

Differentiate $f(x) = e^{5x} + \ln(x)$.

1. Differentiate $e^{5x}$: The derivative of the power ($5x$) is 5. Result: $5e^{5x}$.
2. Differentiate $\ln(x)$: Result: $\frac{1}{x}$.
3. Final Answer: $f'(x) = 5e^{5x} + \frac{1}{x}$.

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4.5 Practice Problems

Part A: Algebra and Log Laws

1. Simplify $2\ln(x) + \ln(y)$.
2. Solve for $x$: $e^{x-3} = 10$.
3. Solve for $x$: $\ln(2x) = 5$.

Part B: Graphs and Features

4. State the horizontal asymptote of $f(x) = e^x + 4$.
5. Find the $x$-intercept of $g(x) = \ln(x – 2)$. (Hint: Set $y=0$ and remember $e^0 = 1$).

Part C: Calculus

6. Find the derivative of $y = 3e^{2x} – 4x^2$.
7. Find the gradient of the curve $f(x) = \ln(x)$ at the point where $x = 5$.
8. Challenge: Find the equation of the tangent line to $f(x) = e^x$ at the point where $x = 0$.

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Solutions (Summary)

• 1. $\ln(x^2 y)$
• 2. $x = \ln(10) + 3 \approx 5.30$
• 3. $x = \frac{e^5}{2} \approx 74.21$
• 4. $y = 4$
• 5. $0 = \ln(x-2) \rightarrow e^0 = x-2 \rightarrow 1 = x-2 \rightarrow x=3$. Point is $(3, 0)$.
• 6. $\frac{dy}{dx} = 6e^{2x} – 8x$
• 7. $f'(x) = \frac{1}{x}$. At $x=5$, gradient is $\frac{1}{5}$.$or 0.2$
• 8. $f'(0) = e^0 = 1$ (gradient). Point is $(0, 1)$. Tangent: $y – 1 = 1(x – 0) \rightarrow y = x + 1$.

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