Exponential and logarithmic functions

IB Mathematics: Analysis and Approaches HL· Unit 2: Functions, Topic 7· 15 min read

1. Definition and Core Properties of Exponential Functions★★☆☆☆⏱ 5 min

📘 Definition

Exponential Function

f(x) = a^x, \quad a > 0, a \neq 1

A function with a constant positive base and variable exponent, defined for all real input values.

Example:

$2^{x}$ , $e^{x}$ , $(\frac{1}{3})^{x}$

All exponential functions share key properties: they are continuous for all real $x$ , they pass through the point $(0, 1)$ (since $a^{0} = 1$ for any $a \neq = 0$ ), and have a horizontal asymptote at $y = 0$ for untransformed functions. If $a > 1$ the function is strictly increasing, if $0 < a < 1$ it is strictly decreasing.

📐 Worked Example

For $f (x) = 3^{(x - 2)} + 1$ , state the domain, range, and equation of the horizontal asymptote.

1
Step 1: The exponent $x - 2$ can take any real value, so horizontal shifts do not change the domain of an exponential function.
2
Domain of $f (x)$ is all real numbers: $(- \infty, \infty)$
3
Step 2: The range of the base exponential $3^{u}$ (for any real $u$ ) is $(0, \infty)$ .
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Step 3: The function is shifted up 1 unit, so the lower bound of the range becomes $0 + 1 = 1$ , with no upper bound.
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Range is $(1, \infty)$
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Step 4: As $x \to - \infty$ , $3^{x - 2} \to 0$ , so $f (x) \to 1$ , so the horizontal asymptote is $y = 1$ .

Exam tip:

Always account for vertical shifts when finding the asymptote and range of transformed exponential functions — this is a common marking point.

2. Logarithms as Inverses and Logarithm Laws★★☆☆☆⏱ 6 min

📘 Definition

Logarithm

\log_a x = y

The inverse of the exponential function $a^{y} = x$ , defined only for positive $x$ , where $a > 0, a \neq = 1$ .

Example:

$lo g_{2} 8 = 3$ , since $2^{3} = 8$

As inverses of exponentials, logarithms swap the domain and range of exponentials: domain is $x > 0$ , range is all real numbers. All core logarithm laws are derived directly from exponent rules.

Product law: $lo g_{a} (x y) = lo g_{a} x + lo g_{a} y$
Quotient law: $lo g_{a} (\frac{x}{y}) = lo g_{a} x - lo g_{a} y$
Power law: $lo g_{a} (x^{k}) = k lo g_{a} x$
Change of base: $lo g_{a} b = \frac{l o g _{c} b}{l o g _{c} a}$ for any $c > 0, c \neq = 1$

📐 Worked Example

Simplify $2 ln 3 + ln 4 - ln 12$ to a single logarithm.

1
Step 1: Apply the power law to the first term:
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$2 \ln 3 = \ln 3^2 = \ln 9$
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Step 2: Apply the product law to add the first two terms:
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$\ln 9 + \ln 4 = \ln (9 \times 4) = \ln 36$
5
Step 3: Apply the quotient law to subtract the final term:
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$\ln 36 - \ln 12 = \ln \left(\frac{36}{12}\right) = \ln 3$

3. Graphs and Transformations★★★☆☆⏱ 5 min

Since exponential and logarithmic functions are inverses, their graphs are reflections of each other over the line $y = x$ . Key features for $a > 1$ (the most common case in exams) are:

Function	Asymptote	Key Point	Monotonicity
$y = a^{x}$	Horizontal $y = 0$	$(0, 1)$	Strictly increasing
$y = lo g_{a} x$	Vertical $x = 0$	$(1, 0)$	Strictly increasing

Transformations follow the same rules as all functions: horizontal shifts move asymptotes horizontally, vertical shifts move them vertically. Always find the new asymptote first before plotting points.

📐 Worked Example

For $y = lo g_{2} (x + 1) - 2$ , state the asymptote and intercepts.

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Step 1: Start with base $y = lo g_{2} x$ , which has vertical asymptote $x = 0$ .
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Step 2: Replace $x$ with $x + 1$ , which shifts the graph left 1 unit. Asymptote becomes $x = - 1$ .
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Step 3: Find the y-intercept by setting $x = 0$ : $y = lo g_{2} (0 + 1) - 2 = 0 - 2 = - 2$ , so y-intercept is $(0, - 2)$ .
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Step 4: Find the x-intercept by setting $y = 0$ :
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$0 = \log_2(x+1) - 2 \implies \log_2(x+1) = 2 \implies x+1 = 2^2 = 4 \implies x=3$
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So x-intercept is $(3, 0)$ .

Exam tip:

Shifted asymptotes are almost always a required marking point in graphing questions for these functions.

4. Solving Exponential and Logarithmic Equations★★★★☆⏱ 7 min

For exponential equations where bases cannot be matched, take logarithms of both sides to bring down the exponent. For logarithmic equations, combine terms into a single logarithm using the laws, then convert to exponential form. Always check for extraneous solutions that fall outside the domain of the original function.

📐 Worked Example

Solve $2 e^{2 x} - 5 e^{x} + 2 = 0$ for $x$ .

1
Step 1: Recognize this is a quadratic equation in $e^{x}$ . Let $u = e^{x}$ , so substitute:
2
$2u^2 - 5u + 2 = 0$
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Step 2: Factor the quadratic:
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$(2u - 1)(u - 2) = 0 \implies u = \frac{1}{2} \text{ or } u = 2$
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Step 3: Substitute back $u = e^{x}$ and take natural logs of both sides:
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$e^x = \frac{1}{2} \implies x = \ln \left(\frac{1}{2}\right) = -\ln 2$
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$e^x = 2 \implies x = \ln 2$
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Step 4: Check: $e^{x}$ is always positive, so both solutions are valid. Final solutions are $x = \pm ln 2$ .

5. Common Pitfalls

Wrong move:

Claiming $lo g_{a} (x + y) = lo g_{a} x + lo g_{a} y$

Why:

Students confuse the product law for logarithms, which applies to products of arguments, not sums inside the logarithm.

Correct move:

Remember $lo g_{a} (x y) = lo g_{a} x + lo g_{a} y$ , this rule does not extend to $lo g_{a} (x + y)$ .

Wrong move:

Forgetting to check for extraneous solutions when solving logarithmic equations

Why:

Algebraic manipulation can produce solutions that are outside the domain of the original logarithmic function.

Correct move:

Substitute all candidate solutions back into the original equation to confirm all logarithm arguments are positive.

Wrong move:

Stating the domain of $a^{x}$ is $x \geq 0$

Why:

Confusing exponential functions (variable exponent) with power functions (variable base), which often have restricted domains.

Correct move:

The domain of any exponential function $a^{x}$ ( $a > 0$ ) is all real numbers $(- \infty, \infty)$ .

Wrong move:

Ignoring that $e^{l n x} = x$ only holds for $x > 0$

Why:

Students assume inverse relations hold for all inputs, but inverses are only defined for inputs in the domain of the inner function.

Correct move:

Any solution for $x \leq 0$ when working with $e^{l n x}$ or $lo g_{a} (a^{x})$ is automatically invalid.

6. Quick Reference Cheatsheet

Feature	Exponential $a^{x}$ , $a > 1$	Logarithm $lo g_{a} x$ , $a > 1$
Domain	$(- \infty, \infty)$	$x > 0$
Range	$y > 0$	$(- \infty, \infty)$
Asymptote	Horizontal: $y = 0$	Vertical: $x = 0$
Key Point	$(0, 1)$	$(1, 0)$
Monotonicity	Strictly increasing	Strictly increasing
Product Rule	$a^{m} a^{n} = a^{m + n}$	$lo g_{a} (x y) = lo g_{a} x + lo g_{a} y$
Quotient Rule	$a^{m} / a^{n} = a^{m - n}$	$lo g_{a} (x / y) = lo g_{a} x - lo g_{a} y$
Power Rule	$(a^{m})^{n} = a^{mn}$	$lo g_{a} x^{k} = k lo g_{a} x$
Inverse Relation	$a^{l o g_{a} x} = x (x > 0)$	$lo g_{a} (a^{x}) = x (\forall x)$

7. Frequently Asked

Do I need to memorize all logarithm laws for the exam?

Yes, all core logarithm laws are expected to be memorized for both Paper 1 (non-calculator) and Paper 2 exams.

Can I use change of base on my calculator for any logarithm?

Yes, most calculators only handle base 10 and natural log, so you are expected to use change of base for other bases in calculator sections.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · Paper 1
Simplify logarithmic expression
2024 · Paper 2
Solve exponential growth problem
2023 · Paper 1
Sketch transformed log graph

Going deeper

What's Next

Exponential and logarithmic functions are foundational for nearly all remaining topics in IB AA HL. They are critical for calculus, where you will learn to differentiate and integrate these functions, and they are the core tool for modeling real-world phenomena like population growth, radioactive decay, and compound interest. They also appear frequently in differential equations, a key topic for Paper 3. Mastering their properties and algebraic manipulation now will pay dividends for more advanced topics later in the course.