IB Mathematics AA – Topic 2: Functions

General Form of Exponential Function:

• \(a\): initial value (y-intercept when \(x=0\))
• \(b\) or \(e^k\): base (determines growth/decay rate)
• If \(b > 1\) or \(k > 0\): exponential growth
• If \(0 < b < 1\) or \(k < 0\): exponential decay
• \(e \approx 2.71828\): natural base (most important in calculus)

Properties of \(f(x) = a \cdot b^x\):

• Domain: All real numbers, \(x \in \mathbb{R}\)
• Range: \(y > 0\) if \(a > 0\); \(y < 0\) if \(a < 0\)
• y-intercept: \((0, a)\) since \(f(0) = a \cdot b^0 = a\)
• x-intercept: None (function never equals zero)
• Horizontal asymptote: \(y = 0\) (x-axis)
• Always increasing if \(b > 1\); always decreasing if \(0 < b < 1\)
• One-to-one: Each output corresponds to exactly one input

General Transformed Form:

\(f(x) = a \cdot b^{k(x-h)} + c\)

• \(a\): vertical stretch/compression and reflection
• \(h\): horizontal shift (right if \(h > 0\))
• \(c\): vertical shift (horizontal asymptote moves to \(y = c\))
• \(k\): horizontal stretch/compression

⚠ Common Pitfalls:

• Confusing \(b^x\) and \(x^b\): Exponential has variable in exponent!
• Domain restriction: \(b\) must be positive (can't take even roots of negatives)
• Base \(b \neq 1\): If \(b = 1\), function is constant, not exponential
• Asymptote confusion: Horizontal asymptote is \(y = c\) (vertical shift), not \(y = 0\) always

Example 1: Analyzing Exponential Functions

Problem: Consider the function \(f(x) = 3 \cdot 2^x - 6\)

(a) State the horizontal asymptote

(b) Find the y-intercept

(c) State the domain and range

(d) Determine if the function represents growth or decay

Solution:

(a) Horizontal asymptote:

The general form is \(f(x) = a \cdot b^x + c\)

Here: \(a = 3\), \(b = 2\), \(c = -6\)

Horizontal asymptote is at \(y = c\)

Horizontal asymptote: \(y = -6\)

(b) y-intercept: Set \(x = 0\)

\(f(0) = 3 \cdot 2^0 - 6 = 3 \cdot 1 - 6 = 3 - 6 = -3\)

y-intercept: \((0, -3)\)

(c) Domain and Range:

Domain: Exponential functions are defined for all real numbers

Domain: \(x \in \mathbb{R}\)

Range: Since \(a = 3 > 0\) and base \(b = 2 > 1\), function grows from asymptote upward

Minimum value approaches asymptote \(y = -6\) (but never reaches it)

Range: \(y > -6\) or \(y \in (-6, \infty)\)

(d) Growth or Decay:

Base \(b = 2 > 1\)

Exponential growth

Logarithm Definition

Key relationship: Logarithmic and exponential functions are inverses

Three Common Logarithms:

• Common logarithm: \(\log_{10}(x)\) often written as \(\log(x)\)
• Natural logarithm: \(\log_e(x)\) written as \(\ln(x)\) (most important in calculus)
• Binary logarithm: \(\log_2(x)\) (used in computer science)

For \(f(x) = \log_b(x)\):

• Domain: \(x > 0\) (only positive numbers have real logarithms)
• Range: All real numbers, \(y \in \mathbb{R}\)
• x-intercept: \((1, 0)\) since \(\log_b(1) = 0\) (any base to power 0 equals 1)
• y-intercept: None (can't evaluate at \(x = 0\))
• Vertical asymptote: \(x = 0\) (y-axis)
• Always increasing if \(b > 1\); always decreasing if \(0 < b < 1\)
• Passes through \((b, 1)\) since \(\log_b(b) = 1\)

Essential Logarithm Laws:

⚠ Logarithm Pitfalls:

• Domain restriction: \(\log(x)\) only defined for \(x > 0\) (no log of zero or negatives in real numbers)
• Not distributive: \(\log(x + y) \neq \log(x) + \log(y)\)
• Power rule direction: \(n\log(x) = \log(x^n)\), not \(\log(nx)\)
• Change of base needed for calculator: Most calculators only have \(\log\) (base 10) and \(\ln\) (base \(e\))

Example 2: Logarithm Laws

Problem: Simplify the following expressions:

(a) \(\log_3(27) + \log_3(9)\)

(b) \(2\ln(x) - \ln(x^2 - 1) + \ln(x + 1)\)

Solution:

(a) \(\log_3(27) + \log_3(9)\)

First, evaluate each logarithm:

\(\log_3(27) = \log_3(3^3) = 3\)

\(\log_3(9) = \log_3(3^2) = 2\)

So: \(3 + 2 = 5\)

Answer: 5

Alternative using product rule:

\(\log_3(27 \times 9) = \log_3(243) = \log_3(3^5) = 5\)

(b) \(2\ln(x) - \ln(x^2 - 1) + \ln(x + 1)\)

Apply power rule: \(2\ln(x) = \ln(x^2)\)

\(= \ln(x^2) - \ln(x^2 - 1) + \ln(x + 1)\)

Apply quotient rule to first two terms:

\(= \ln\left(\frac{x^2}{x^2 - 1}\right) + \ln(x + 1)\)

Apply product rule:

\(= \ln\left(\frac{x^2(x + 1)}{x^2 - 1}\right)\)

Factor denominator: \(x^2 - 1 = (x-1)(x+1)\)

\(= \ln\left(\frac{x^2(x + 1)}{(x-1)(x+1)}\right)\)

Cancel \((x+1)\):

\(= \ln\left(\frac{x^2}{x-1}\right)\)

Three Methods:

Strategy:

1. Combine logarithms using log laws (if multiple logs)
2. Exponentiate both sides to eliminate logarithm
3. Solve the resulting equation
4. Check domain: Verify solution makes arguments positive

Example: \(\log_2(x) + \log_2(x-3) = 2\)

\(\log_2(x(x-3)) = 2\)

\(x(x-3) = 2^2 = 4\)

\(x^2 - 3x - 4 = 0\)

\((x-4)(x+1) = 0\)

\(x = 4\) or \(x = -1\)

Check: \(x = -1\) gives \(\log_2(-1)\) which is undefined

Solution: \(x = 4\) only

💡 Solving Tips:

• For exponential equations: isolate exponential term, then take logarithms
• For logarithmic equations: combine logs, then exponentiate
• Always check solutions in original equation (especially for logs)
• Use \(\ln\) when base is \(e\), otherwise use change of base
• GDC can solve numerically—use for verification or when algebra is complex

Example 3: Solving Exponential and Logarithmic Equations (IB-Style)

Problem:

(a) Solve: \(5^{2x-1} = 200\) (give answer to 3 significant figures)

(b) Solve: \(\ln(x+2) + \ln(x-1) = \ln(8)\)

Solution:

(a) \(5^{2x-1} = 200\)

Take logarithm of both sides (using any base—here we'll use common log):

\(\log(5^{2x-1}) = \log(200)\)

Apply power rule:

\((2x-1)\log(5) = \log(200)\)

Divide both sides by \(\log(5)\):

\(2x - 1 = \frac{\log(200)}{\log(5)}\)

\(2x = \frac{\log(200)}{\log(5)} + 1\)

\(2x = \frac{2.301}{0.699} + 1\)

\(2x = 3.292 + 1 = 4.292\)

\(x = 2.146\)

\(x = 2.15\) (to 3 s.f.)

(b) \(\ln(x+2) + \ln(x-1) = \ln(8)\)

Apply product rule on left side:

\(\ln[(x+2)(x-1)] = \ln(8)\)

Since logs are equal, arguments must be equal:

\((x+2)(x-1) = 8\)

Expand:

\(x^2 + x - 2 = 8\)

\(x^2 + x - 10 = 0\)

Use quadratic formula:

\(x = \frac{-1 \pm \sqrt{1 + 40}}{2} = \frac{-1 \pm \sqrt{41}}{2}\)

\(x = \frac{-1 + 6.403}{2} = 2.702\) or \(x = \frac{-1 - 6.403}{2} = -3.702\)

Check domains:

For \(x = 2.702\): \(x + 2 = 4.702 > 0\) ✓ and \(x - 1 = 1.702 > 0\) ✓

For \(x = -3.702\): \(x + 2 = -1.702 < 0\) ✗ (can't take log of negative)

\(x = \frac{-1 + \sqrt{41}}{2} \approx 2.70\)

Key Graph Features:

Property	Exponential \(y = b^x\)	Logarithmic \(y = \log_b(x)\)
Domain	All reals (\(\mathbb{R}\))	\(x > 0\)
Range	\(y > 0\)	All reals (\(\mathbb{R}\))
Asymptote	Horizontal: \(y = 0\)	Vertical: \(x = 0\)
x-intercept	None	\((1, 0)\)
y-intercept	\((0, 1)\)	None
Key Point	\((1, b)\)	\((b, 1)\)
Relationship	Inverse functions—reflect across \(y = x\)

Steps for Sketching:

1. Identify the function type (exponential or logarithmic)
2. Find and draw asymptotes (dashed lines)
3. Find and plot intercepts
4. Identify key points (like \((1, b)\) for exponential, \((b, 1)\) for log)
5. Determine if function is increasing or decreasing
6. Sketch smooth curve through points, approaching asymptotes
7. Verify with GDC

⚠ Graphing Mistakes:

• Wrong asymptote direction: Exponential has horizontal, logarithmic has vertical
• Crossing asymptotes: Curves approach but never touch/cross asymptotes
• Wrong domain: Logarithmic only defined for \(x > 0\)
• Incorrect intercepts: Know which functions have which intercepts

Essential Formulas:

• \(y = \log_b(x) \Leftrightarrow b^y = x\)
• \(\log_b(xy) = \log_b(x) + \log_b(y)\)
• \(\log_b(x/y) = \log_b(x) - \log_b(y)\)
• \(\log_b(x^n) = n\log_b(x)\)
• Change of base: \(\log_b(x) = \frac{\ln(x)}{\ln(b)}\)
• \(e^{\ln(x)} = x\) and \(\ln(e^x) = x\)

Common Question Types:

• "Solve exponentially": Take logs of both sides or match bases
• "Simplify using log laws": Apply product, quotient, power rules
• "Sketch the function": Identify asymptote, intercepts, key points
• "Find the inverse": Exponential and log are inverses—swap and solve
• "Express in form \(y = ae^{kx} + c\)": Use log laws to rearrange

Calculator Tips:

• Use \(\ln\) button for natural log, \(\log\) for base 10
• For other bases, use change of base formula
• Graph to verify solutions and check domain
• Use solver/equation mode for numerical solutions