Module 0e — Exponential and Logarithmic Functions
Course: MAT1CJ101 — Differential Calculus (prerequisite refresher)
Status: Prerequisite background material — not part of the official 60 taught hours, not examined.
0. Why This Module Exists
The derivative of $e^x$ and $\ln x$ are among the first “new” derivative rules you meet after the power rule, and they only make sense if you’re fluent in the algebra of exponents and logs first — otherwise every derivative computation doubles as an algebra lesson. This module is that algebra, isolated and drilled before calculus needs it.
1. Laws of Exponents
For $a, b > 0$ and any real exponents $m, n$:
| Law | Statement |
|---|---|
| Product | $a^m \cdot a^n = a^{m+n}$ |
| Quotient | $a^m / a^n = a^{m-n}$ |
| Power of a power | $(a^m)^n = a^{mn}$ |
| Power of a product | $(ab)^n = a^n b^n$ |
| Power of a quotient | $(a/b)^n = a^n/b^n$ |
| Zero exponent | $a^0 = 1$ ($a \neq 0$) |
| Negative exponent | $a^{-n} = 1/a^n$ |
| Fractional exponent | $a^{1/n} = \sqrt[n]{a}$, $\ a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$ |
Worked Example 1: Simplify $(2x^3y^{-2})^4 / (4x^{-1}y^2)$.
\(\text{Numerator: } (2x^3y^{-2})^4 = 2^4 x^{3 \cdot 4} y^{-2 \cdot 4} = 16 x^{12} y^{-8}\) \(\text{Full expression: } \frac{16x^{12}y^{-8}}{4x^{-1}y^2} = \frac{16}{4} \cdot x^{12-(-1)} \cdot y^{-8-2} = 4x^{13}y^{-10} = \frac{4x^{13}}{y^{10}}\)
Worked Example 2: Simplify $8^{2/3}$.
\[8^{2/3} = (\sqrt[3]{8})^2 \quad \text{[denominator 3} \to \text{cube root, numerator 2} \to \text{then square]} = 2^2 = 4\]2. Exponential Functions
2.1 Definition and Graph Shape
An exponential function has the form $f(x) = a \cdot b^x$, where $b > 0$, $b \neq 1$ (the base), and $a \neq 0$.
| Feature | $b > 1$ (growth) | $0 < b < 1$ (decay) |
|---|---|---|
| Behaviour as $x$ increases | Increases | Decreases |
| As $x \to \infty$ | $\to \infty$ | $\to 0$ |
| As $x \to -\infty$ | $\to 0$ | $\to \infty$ |
| $y$-intercept | $(0, a)$ always, since $b^0 = 1$ | $(0, a)$ always |
| Horizontal asymptote | $y = 0$ | $y = 0$ |
Unlike a polynomial, an exponential function is never zero ($b^x > 0$ for all real $x$ when $b > 0$), so the $x$-axis is a horizontal asymptote it approaches but never touches.
Worked Example 3: Compare $f(x) = 2^x$ and $g(x) = (1/2)^x$ at $x = -2, 0, 3$.
\(f(-2) = 2^{-2} = 1/4 \qquad g(-2) = (1/2)^{-2} = 2^2 = 4\) \(f(0) = 2^0 = 1 \qquad g(0) = (1/2)^0 = 1\) \(f(3) = 2^3 = 8 \qquad g(3) = (1/2)^3 = 1/8\)
Notice: $g(x) = (1/2)^x = 2^{-x} = f(-x)$. $g$ is $f$ reflected across the $y$-axis — makes sense, since base $1/2$ is the reciprocal of base 2.
2.2 The Number e
$e \approx 2.71828\ldots$ is the base that makes the exponential function its own derivative — the unique real number for which the graph of $e^x$ has slope exactly 1 at $x = 0$. You’ll meet $e$ formally as a limit, $\lim_{n \to \infty} (1 + 1/n)^n$, in Module I/II; for now, treat it as “the natural base” for exponentials, the way $\pi$ is the natural constant for circles.
3. Logarithmic Functions
3.1 Logarithms as Inverses
Definition: for $b > 0$, $b \neq 1$, $\log_b(x)$ is defined by:
\[y = \log_b(x) \iff b^y = x \quad (x > 0)\]The logarithm base $b$ is precisely the inverse function of $b^x$ (Module 0b, Section 4) — this is why $\log_b(x)$ is only defined for $x > 0$: the range of $b^x$ is $(0, \infty)$, and the domain of an inverse function is the range of the original.
| Base | Notation | Name |
|---|---|---|
| 10 | $\log(x)$ or $\log_{10}(x)$ | Common logarithm |
| $e$ | $\ln(x)$ | Natural logarithm |
| $b$ (general) | $\log_b(x)$ | Logarithm base $b$ |
Worked Example 4: Convert between exponential and logarithmic form.
\(2^3 = 8 \iff \log_2(8) = 3\) \(10^{-2} = 0.01 \iff \log_{10}(0.01) = -2\) \(e^0 = 1 \iff \ln(1) = 0\) \(5^1 = 5 \iff \log_5(5) = 1\)
3.2 Basic Log Values
Because $\log_b$ is the inverse of $b^x$, these follow directly:
| Fact | Reason |
|---|---|
| $\log_b(1) = 0$ | $b^0 = 1$ |
| $\log_b(b) = 1$ | $b^1 = b$ |
| $\log_b(b^x) = x$ | inverse cancels the base |
| $b^{\log_b x} = x$ | inverse cancels the log |
3.3 Laws of Logarithms
For $x, y > 0$, $b > 0$, $b \neq 1$, and any real $r$:
| Law | Statement |
|---|---|
| Product | $\log_b(xy) = \log_b(x) + \log_b(y)$ |
| Quotient | $\log_b(x/y) = \log_b(x) - \log_b(y)$ |
| Power | $\log_b(x^r) = r \cdot \log_b(x)$ |
| Change of base | $\log_b(x) = \ln(x)/\ln(b) = \log(x)/\log(b)$ |
Why the product law is true (not just a rule to memorise): let $m = \log_b(x)$, $n = \log_b(y)$, so $x = b^m$, $y = b^n$. Then $xy = b^m \cdot b^n = b^{m+n}$, so $\log_b(xy) = m + n = \log_b(x) + \log_b(y)$. Every log law is really an exponent law in disguise.
Worked Example 5: Expand $\log_2(8x^3/y)$ fully using the log laws.
\(\log_2(8x^3/y) = \log_2(8x^3) - \log_2(y) \quad \text{[quotient law]}\) \(= \log_2(8) + \log_2(x^3) - \log_2(y) \quad \text{[product law]}\) \(= \log_2(8) + 3\log_2(x) - \log_2(y) \quad \text{[power law]}\) \(= 3 + 3\log_2(x) - \log_2(y) \quad \text{[}\log_2(8) = \log_2(2^3) = 3\text{]}\)
Worked Example 6: Combine $2\ln(x) + \ln(y) - 3\ln(z)$ into a single logarithm.
\(2\ln(x) = \ln(x^2) \quad \text{[power law, reversed]}\) \(\ln(y) \text{ stays as is}\) \(3\ln(z) = \ln(z^3) \quad \text{[power law, reversed]}\) \(\ln(x^2) + \ln(y) - \ln(z^3) = \ln(x^2y) - \ln(z^3) \quad \text{[product law, reversed]}\) \(= \ln\!\left(\frac{x^2y}{z^3}\right) \quad \text{[quotient law, reversed]}\)
3.4 Change of Base
Calculators typically only compute $\log_{10}$ or $\ln$ directly. To find $\log_b(x)$ for another base, use the change of base formula: $\log_b(x) = \ln(x)/\ln(b)$.
Worked Example 7: Evaluate $\log_3(20)$ to 4 decimal places.
\[\log_3(20) = \frac{\ln(20)}{\ln(3)} \approx \frac{2.9957}{1.0986} \approx 2.7268\]4. Solving Exponential and Logarithmic Equations
4.1 Exponential Equations — Same Base Method
If bases can be matched, equate the exponents directly (since $b^x$ is injective — Module 0b — equal outputs force equal inputs).
Worked Example 8: Solve $3^{2x-1} = 27$.
Rewrite 27 as a power of 3: $27 = 3^3$
\[3^{2x-1} = 3^3\]Since $b^x$ is injective, exponents must match: \(2x - 1 = 3 \implies 2x = 4 \implies x = 2\)
4.2 Exponential Equations — Taking Logarithms
When bases can’t be matched, take a logarithm of both sides.
Worked Example 9: Solve $5^x = 40$.
Take $\ln$ of both sides: \(\ln(5^x) = \ln(40) \implies x \cdot \ln(5) = \ln(40) \quad \text{[power law]}\) \(x = \frac{\ln(40)}{\ln(5)} \approx \frac{3.6889}{1.6094} \approx 2.2920\)
Worked Example 10: Solve $2 \cdot 3^{x+1} = 50$.
\(3^{x+1} = 25 \quad \text{[divide both sides by 2]}\) \(\ln(3^{x+1}) = \ln(25) \implies (x+1) \cdot \ln(3) = \ln(25)\) \(x + 1 = \frac{\ln(25)}{\ln(3)} \approx \frac{3.2189}{1.0986} \approx 2.9299\) \(x \approx 1.9299\)
4.3 Logarithmic Equations
Convert to exponential form, or combine logs first using the laws, then convert.
Worked Example 11: Solve $\log_2(x + 3) = 4$.
Convert to exponential form: $x + 3 = 2^4 = 16 \implies x = 13$
Domain check: need $x+3 > 0$, i.e. $x > -3$. $x=13$ satisfies this. $\checkmark$
Worked Example 12: Solve $\log(x) + \log(x - 3) = 1$ (base 10).
Combine using product law: $\log[x(x-3)] = 1$
Convert to exponential form: $x(x-3) = 10^1 = 10$ \(x^2 - 3x - 10 = 0 \implies (x-5)(x+2) = 0 \implies x = 5 \text{ or } x = -2\)
Domain check: need $x > 0$ AND $x - 3 > 0$, i.e. $x > 3$.
$x = 5$: $5 > 3$ $\checkmark$ valid
$x = -2$: fails $x > 3$ (and fails $x > 0$) $\times$ REJECTED (extraneous)
Solution: $x = 5$ only.
Critical habit: always check candidate solutions against the domain of the original logarithmic expression — combining logs can introduce extraneous roots, exactly as squaring both sides of an equation can.
5. Summary
| Concept | Key Idea | Worked Example |
|---|---|---|
| Laws of exponents | $a^ma^n=a^{m+n}$, $(a^m)^n=a^{mn}$, etc. | $(2x^3y^{-2})^4/(4x^{-1}y^2) = 4x^{13}/y^{10}$ |
| Exponential function | $f(x)=a\cdot b^x$, never zero, asymptote $y=0$ | $2^x$ vs $(1/2)^x$ are mirror images |
| $e$ | Natural base, slope 1 at $x=0$ | — |
| $\log_b(x)$ | Inverse of $b^x$; defined for $x>0$ | $2^3=8 \iff \log_2 8=3$ |
| Log laws | $\log(xy)=\log x+\log y$, etc. — from exponent laws | $\log_2(8x^3/y)$ expanded |
| Change of base | $\log_b(x)=\ln(x)/\ln(b)$ | $\log_3(20)\approx2.7268$ |
| Exponential equation, same base | Equate exponents (injectivity) | $3^{2x-1}=27 \implies x=2$ |
| Exponential equation, general | Take $\ln$ of both sides | $5^x=40 \implies x\approx2.292$ |
| Log equation | Convert to exponential form; check domain | $\log(x)+\log(x-3)=1 \implies x=5$ (only) |
6. Practice Problems
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Simplify $(3a^2b^{-1})^3 / (9a^{-1}b^2)$.
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Simplify $16^{3/4}$.
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Simplify $(x^{2/3} \cdot x^{1/6}) / x^{1/2}$.
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Sketch (describe key features: intercept, asymptote, increasing/decreasing) $f(x) = 3^x$ and $g(x) = 3^{-x}$ on the same axes.
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Convert to logarithmic form: (a) $4^3 = 64$, (b) $10^{-1} = 0.1$, (c) $e^2 = x$.
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Convert to exponential form: (a) $\log_5(125) = 3$, (b) $\ln(x) = 2$, (c) $\log(0.001) = -3$.
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Evaluate without a calculator: $\log_2(32)$, $\log_3(1/9)$, $\log_7(1)$, $\ln(e^5)$.
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Expand fully using log laws: $\log_5(25x^2/y^3)$.
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Combine into a single logarithm: $3\ln(x) - 2\ln(y) + \ln(z)$.
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Use change of base to evaluate $\log_7(50)$ to 4 decimal places.
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Solve $4^{3x} = 64$.
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Solve $2^{x-1} = 3^x$. (Hint: you can’t match bases here — take $\ln$ of both sides and solve for $x$, isolating $x$ on one side.)
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Solve $7 \cdot 2^x = 56$.
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Solve $\log_3(2x - 1) = 2$.
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Solve $\log(x + 2) + \log(x - 1) = 1$. Check for extraneous solutions.
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Solve $\ln(x^2) - \ln(x + 6) = 0$. Check for extraneous solutions.
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(Harder) Solve $2^{2x} - 5 \cdot 2^x + 4 = 0$. (Hint: let $u = 2^x$, solve the resulting quadratic in $u$, then recover $x$ for each valid value of $u$ — reject any $u \leq 0$.)
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(Harder) A quantity grows according to $A(t) = A_0 e^{kt}$. If $A(0) = 200$ and $A(5) = 500$, find $k$ (in terms of natural logs) and then find $A(10)$.