Chapter 6: Logarithmic and Exponential Functions
6.1 Exponential Functions
Definition:
An exponential function is a function of the form:
\[
f(x) = a^x \quad \text{where } a > 0, a \neq 1
\]
Special case: The natural exponential function:
\[
f(x) = e^x
\]
where \(e \approx 2.71828\ldots\) (Euler's number).
Key properties of \(f(x) = a^x\):
| Property |
Description |
| Domain |
All real numbers (\(\mathbb{R}\)) |
| Range |
\(y > 0\) (positive only) |
| y-intercept |
\((0, 1)\) since \(a^0 = 1\) |
| Asymptote |
x-axis (\(y = 0\)) as \(x \to -\infty\) (for \(a > 1\)) |
| Increasing |
If \(a > 1\), function is increasing |
| Decreasing |
If \(0 < a < 1\), function is decreasing |
Proof: \(a^x\) is always positive for \(a > 0\)
For \(x > 0\): \(a^x > 0\) (obvious).
For \(x = 0\): \(a^0 = 1 > 0\).
For \(x < 0\): \(a^x = \dfrac{1}{a^{-x}}\). Since \(a^{-x} > 0\), the reciprocal is also positive.
Thus \(a^x > 0\) for all real \(x\).
Worked Example 6.1 (Exponential graphs – non-calculator)
Sketch the graph of \(y = 2^x\) for \(-3 \le x \le 3\). Label key points.
Solution:
| \(x\) |
-3 |
-2 |
-1 |
0 |
1 |
2 |
3 |
| \(y\) |
\(\frac{1}{8}\) |
\(\frac{1}{4}\) |
\(\frac{1}{2}\) |
1 |
2 |
4 |
8 |
Shape: increasing curve, passes through \((0,1)\), approaches x-axis as \(x \to -\infty\), rises steeply for \(x > 0\).
Worked Example 6.2 (Mauritian context – population growth)
The population of a town in Mauritius grows according to \(P(t) = 1000(1.05)^t\), where \(t\) is the number of years since 2020. Find the population in 2025.
Solution:
\(t = 2025 - 2020 = 5\)
\(P(5) = 1000(1.05)^5 \approx 1000 \times 1.2763 = 1276\)
Answer: Approximately 1276 people.
6.2 Logarithmic Functions
Definition:
The logarithm of a number \(y\) to base \(a\) is the exponent to which \(a\) must be raised to get \(y\):
\[
\log_a y = x \quad \iff \quad a^x = y
\]
Special cases:
- Common logarithm: \(\log_{10} x\) or \(\log x\) (base 10)
- Natural logarithm: \(\ln x\) (base \(e\))
Key properties of \(f(x) = \log_a x\):
| Property |
Description |
| Domain |
\(x > 0\) (positive only) |
| Range |
All real numbers (\(\mathbb{R}\)) |
| x-intercept |
\((1, 0)\) since \(\log_a 1 = 0\) |
| Asymptote |
y-axis (\(x = 0\)) as \(x \to 0^+\) |
| Increasing |
If \(a > 1\), function is increasing |
| Decreasing |
If \(0 < a < 1\), function is decreasing |
Proof: Exponential and logarithmic functions are inverses
Given \(f(x) = a^x\) and \(g(x) = \log_a x\):
Domain of \(f\) = \(\mathbb{R}\), range = \((0, \infty)\)
Domain of \(g\) = \((0, \infty)\), range = \(\mathbb{R}\)
Composition:
\[
f(g(x)) = a^{\log_a x} = x \quad \text{for } x > 0
\]
\[
g(f(x)) = \log_a (a^x) = x \quad \text{for all } x \in \mathbb{R}
\]
This satisfies the definition of inverse functions.
Graphical implication:
The graphs of \(y = a^x\) and \(y = \log_a x\) are reflections of each other across the line \(y = x\).
Worked Example 6.3 (Converting between exponential and logarithmic form)
Write each in the other form:
- \(3^4 = 81\)
- \(\log_2 8 = 3\)
Solution:
(a) Exponential form: \(3^4 = 81\) \(\rightarrow\) Logarithmic form: \(\log_3 81 = 4\)
(b) Logarithmic form: \(\log_2 8 = 3\) \(\rightarrow\) Exponential form: \(2^3 = 8\)
Worked Example 6.4 (Evaluating logarithms – non-calculator)
Evaluate:
- \(\log_2 16\)
- \(\log_5 125\)
- \(\log_3 \frac{1}{9}\)
Solution:
(a) \(2^4 = 16\) \(\rightarrow\) \(\log_2 16 = 4\)
(b) \(5^3 = 125\) \(\rightarrow\) \(\log_5 125 = 3\)
(c) \(3^{-2} = \frac{1}{9}\) \(\rightarrow\) \(\log_3 \frac{1}{9} = -2\)
6.3 Laws of Logarithms
Law 1 (Product):
\[
\log_a (xy) = \log_a x + \log_a y
\]
Law 2 (Quotient):
\[
\log_a \left(\frac{x}{y}\right) = \log_a x - \log_a y
\]
Law 3 (Power):
\[
\log_a (x^n) = n \log_a x
\]
Law 4 (Change of Base):
\[
\log_a x = \frac{\log_b x}{\log_b a}
\]
Proof: Law 1 (Product)
Let \(p = \log_a x\) and \(q = \log_a y\). Then \(x = a^p\) and \(y = a^q\).
Then \(xy = a^p a^q = a^{p+q}\).
So \(\log_a (xy) = p + q = \log_a x + \log_a y\).
Proof: Law 2 (Quotient)
Let \(p = \log_a x\) and \(q = \log_a y\). Then \(x = a^p\) and \(y = a^q\).
Then \(\frac{x}{y} = \frac{a^p}{a^q} = a^{p-q}\).
So \(\log_a \left(\frac{x}{y}\right) = p - q = \log_a x - \log_a y\).
Proof: Law 3 (Power)
Let \(p = \log_a x\). Then \(x = a^p\).
Then \(x^n = (a^p)^n = a^{np}\).
So \(\log_a (x^n) = np = n \log_a x\).
Proof: Law 4 (Change of Base)
Let \(y = \log_a x\). Then \(a^y = x\).
Take \(\log_b\) of both sides: \(\log_b (a^y) = \log_b x\).
Using Law 3: \(y \log_b a = \log_b x\).
Thus \(y = \frac{\log_b x}{\log_b a}\).
Worked Example 6.5 (Simplifying logarithms – non-calculator)
Write as a single logarithm:
- \(\log_2 8 + \log_2 4\)
- \(3 \log_5 2 + \log_5 3\)
Solution:
(a) \(\log_2 8 + \log_2 4 = \log_2 (8 \times 4) = \log_2 32 = 5\)
(b) \(3 \log_5 2 + \log_5 3 = \log_5 (2^3) + \log_5 3 = \log_5 (8 \times 3) = \log_5 24\)
Worked Example 6.6 (Using change of base)
Evaluate \(\log_4 8\) without a calculator.
Solution:
\[
\log_4 8 = \frac{\log_2 8}{\log_2 4} = \frac{3}{2}
\]
Answer: \(\frac{3}{2}\)
Worked Example 6.7 (Mauritian context – pH scale)
The pH of a solution is given by \(\text{pH} = -\log_{10} [H^+]\), where \([H^+]\) is the hydrogen ion concentration. If the pH of a river in Mauritius is 6.5, find \([H^+]\).
Solution:
\[
6.5 = -\log_{10} [H^+] \Rightarrow \log_{10} [H^+] = -6.5
\]
\[
[H^+] = 10^{-6.5} \approx 3.16 \times 10^{-7}
\]
Answer: \([H^+] \approx 3.16 \times 10^{-7}\) moles per litre.
6.4 Solving Exponential Equations \(a^x = b\)
To solve \(a^x = b\):
- Take logs of both sides (usually base 10 or base \(e\))
- Use the power law: \(\log(a^x) = x \log a\)
- Solve for \(x\): \(x = \frac{\log b}{\log a}\)
Proof: Taking logs preserves equality
If two positive quantities are equal, their logarithms (to any base) are equal.
So \(a^x = b \Rightarrow \log(a^x) = \log b \Rightarrow x \log a = \log b \Rightarrow x = \frac{\log b}{\log a}\).
Worked Example 6.8 (Solving exponential equation – non-calculator)
Solve \(2^x = 32\).
Solution:
\(2^x = 2^5 \Rightarrow x = 5\)
Answer: \(x = 5\)
Worked Example 6.9 (Solving exponential equation – calculator required)
Solve \(3^{2x-1} = 7\).
Solution:
\[
\log_{10} (3^{2x-1}) = \log_{10} 7
\]
\[
(2x - 1) \log_{10} 3 = \log_{10} 7
\]
\[
2x - 1 = \frac{\log_{10} 7}{\log_{10} 3}
\]
\[
2x = 1 + \frac{\log_{10} 7}{\log_{10} 3}
\]
\[
x = \frac{1 + \frac{\log_{10} 7}{\log_{10} 3}}{2} \approx \frac{1 + 1.7712}{2} = 1.3856
\]
Answer: \(x \approx 1.386\)
Worked Example 6.10 (Mauritian context – compound interest)
A savings account in a Mauritian bank offers 4% interest per year, compounded annually. How long will it take for an investment of Rs 10,000 to grow to Rs 15,000?
Solution:
\[
10000(1.04)^t = 15000
\]
\[
(1.04)^t = 1.5
\]
\[
t = \frac{\log_{10} 1.5}{\log_{10} 1.04} \approx \frac{0.1761}{0.0170} \approx 10.36
\]
Answer: About 10.4 years.
6.5 Graphs of \(k e^{nx} + a\) and \(k \ln(ax + b)\)
For \(y = k e^{nx} + a\):
| Feature |
Description |
| Shape |
Exponential growth (if \(n > 0\)) or decay (if \(n < 0\)) |
| Horizontal asymptote |
\(y = a\) |
| y-intercept |
\(x = 0\): \(y = k e^0 + a = k + a\) |
| Increasing |
If \(kn > 0\), increasing; if \(kn < 0\), decreasing |
For \(y = k \ln(ax + b)\):
| Feature |
Description |
| Shape |
Logarithmic (gradual increase) |
| Vertical asymptote |
Solve \(ax + b = 0 \Rightarrow x = -\frac{b}{a}\) |
| x-intercept |
Solve \(k \ln(ax + b) = 0 \Rightarrow ax + b = 1\) |
| Domain |
\(ax + b > 0\) |
Worked Example 6.11 (Sketching \(k e^{nx} + a\))
Sketch \(y = 2e^{0.5x} - 3\). Label the asymptote and y-intercept.
Solution:
- Asymptote: \(y = -3\)
- y-intercept: \(x = 0 \Rightarrow y = 2e^0 - 3 = 2 - 3 = -1\)
- Since \(k = 2 > 0\) and \(n = 0.5 > 0\), the graph is increasing
Sketch: exponential growth curve starting at \((0, -1)\), approaching \(y = -3\) as \(x \to -\infty\), rising steeply for \(x > 0\).
Worked Example 6.12 (Sketching \(k \ln(ax + b)\))
Sketch \(y = 3 \ln(x - 2) + 1\). Label the asymptote and x-intercept.
Solution:
- Asymptote: \(x - 2 = 0 \Rightarrow x = 2\)
- x-intercept: \(3 \ln(x - 2) + 1 = 0 \Rightarrow \ln(x - 2) = -\frac{1}{3} \Rightarrow x - 2 = e^{-1/3} \Rightarrow x = 2 + e^{-1/3} \approx 2.716\)
- Domain: \(x > 2\)
Sketch: logarithmic curve starting just above \(x = 2\), crossing x-axis at \(x \approx 2.716\), gradually increasing.
Chapter 6 Summary
| Concept |
Key points |
| Exponential |
\(f(x) = a^x\), \(a > 0, a \neq 1\) |
| Logarithm |
\(\log_a y = x \iff a^x = y\) |
| Inverse relationship |
\(a^x\) and \(\log_a x\) are inverses |
| Laws of logs |
Product, quotient, power, change of base |
| Solving \(a^x = b\) |
\(x = \frac{\log b}{\log a}\) |
| Graphs |
Exponential has horizontal asymptote; logarithmic has vertical asymptote |
Exercises – Chapter 6
Easy (Drill – Non-Calculator)
- Evaluate without a calculator:
(a) \(\log_2 32\)
(b) \(\log_3 81\)
(c) \(\log_5 \frac{1}{25}\)
- Write as a single logarithm:
(a) \(\log_3 5 + \log_3 7\)
(b) \(2 \log_4 3 - \log_4 9\)
- Sketch \(y = 3^x\) and \(y = \log_3 x\) on the same axes for \(0 \le x \le 4\).
Medium (Examination Style)
- Solve for \(x\):
(a) \(5^x = 125\)
(b) \(2^{x+1} = 32\)
(c) \(3^{2x-1} = 27\)
- Simplify: \(\log_2 16 + \log_2 8 - \log_2 4\)
- Mauritian context: The number of tourists visiting Mauritius since 2010 can be modelled by \(N(t) = 800000 e^{0.05t}\), where \(t\) is the number of years after 2010. Find the number of tourists in 2020.
- Solve \(2^x = 10\), giving your answer to 3 significant figures.
Hard (Challenge for A*)
- Solve \(\log_2 x + \log_2 (x - 2) = 3\).
- Prove that \(\log_a b = \frac{1}{\log_b a}\) for \(a, b > 0, a \neq 1, b \neq 1\).
- Mauritian context: A population of deer in a national park is given by:
\[
P(t) = 500 e^{-0.02t} + 200
\]
where \(t\) is the number of years since 2020.
(a) What is the initial population (in 2020)?
(b) What is the population after 10 years?
(c) What is the limiting population as \(t \to \infty\)?
(d) Sketch the graph, labelling the asymptote and y-intercept.