The Exponential Function Family $y = a^x$ and $y = e^x$
- Parent Exponential Functions: For base $a > 0, a \ne 1$ (and specifically natural base $e \approx 2.71828$):
- Domain: $\mathbb{R}$ (all real numbers).
- Range: $\{y \in \mathbb{R} \mid y > 0\}$ (strictly positive).
- Horizontal Asymptote: $y = 0$ ($x$-axis). For $a > 1$, as $x \to -\infty$, $a^x \to 0$.
- $y$-intercept: $(0, 1)$ because $a^0 = 1$.
- No $x$-intercepts ($a^x = 0$ has no real solution).
- Growth vs Decay: If $a > 1$, the function is strictly increasing (exponential growth); if $0 < a < 1$, it is strictly decreasing (exponential decay).
- Transformed Exponential Form: $f(x) = A \cdot b^{x - h} + k$
- Horizontal asymptote shifts to $y = k$.
- If $A > 0$, range is $\{y \in \mathbb{R} \mid y > k\}$; if $A < 0$, range is $\{y \in \mathbb{R} \mid y < k\}$.
(a) State the equation of the horizontal asymptote and the range of $f$.
(b) Find the coordinates of the $y$-intercept and $x$-intercept.
Part (a): Asymptote & Range
Horizontal asymptote: $y = -6$.
Since $A = 3 > 0$, the range is $\{y \in \mathbb{R} \mid y > -6\}$ (or $(-6, \infty)$).
Part (b): Axial Intercepts
$y$-intercept (set $x = 0$):
$$f(0) = 3 \cdot 2^{0 - 1} - 6 = 3 \cdot 2^{-1} - 6 = \frac{3}{2} - 6 = -4.5 \implies (0, -4.5)$$$x$-intercept (set $f(x) = 0$):
$$3 \cdot 2^{x - 1} - 6 = 0 \implies 3 \cdot 2^{x - 1} = 6 \implies 2^{x - 1} = 2^1$$ $$x - 1 = 1 \implies x = 2 \implies (2, 0)$$The Logarithmic Function Family $y = \log_a x$ and $y = \ln x$
- Parent Logarithmic Functions:
- Domain: $\{x \in \mathbb{R} \mid x > 0\}$ (the argument must be strictly positive!).
- Range: $\mathbb{R}$ (all real numbers).
- Vertical Asymptote: $x = 0$ ($y$-axis). As $x \to 0^+$, $\ln x \to -\infty$.
- $x$-intercept: $(1, 0)$ because $\ln 1 = 0$.
- No $y$-intercept ($x = 0$ is strictly outside the domain).
- Transformed Logarithmic Form: $g(x) = A \ln(b(x - h)) + k$
- Vertical asymptote occurs where argument $= 0 \implies x = h$.
- Domain condition: $b(x - h) > 0$.
(a) State the domain and the equation of the vertical asymptote.
(b) Find the exact coordinates of the $x$-intercept.
Part (a): Domain & Asymptote
Condition: argument must be strictly positive:
$$2x - 8 > 0 \implies 2x > 8 \implies x > 4$$Domain: $\{x \in \mathbb{R} \mid x > 4\}$. Vertical asymptote: $x = 4$.
Part (b): $x$-intercept (set $g(x) = 0$)
$$\ln(2x - 8) - 1 = 0 \implies \ln(2x - 8) = 1$$ $$2x - 8 = e^1 = e \implies 2x = e + 8 \implies x = \frac{e + 8}{2}$$Answer: $\left(\frac{e + 8}{2}, 0\right)$ (approximately $(5.36, 0)$).
Inverse Duality & Reflection Across $y = x$
- Mutual Inverses: $$e^{\ln x} = x \quad (x > 0) \qquad \text{and} \qquad \ln(e^x) = x \quad (x \in \mathbb{R})$$
- Feature Exchange Under Inversion:
- Horizontal asymptote $y = 0$ of $e^x \longleftrightarrow$ Vertical asymptote $x = 0$ of $\ln x$.
- $y$-intercept $(0, 1)$ of $e^x \longleftrightarrow$ $x$-intercept $(1, 0)$ of $\ln x$.
- Domain $\mathbb{R}$ of $e^x \longleftrightarrow$ Range $\mathbb{R}$ of $\ln x$.
- Range $(0, \infty)$ of $e^x \longleftrightarrow$ Domain $(0, \infty)$ of $\ln x$.
Step 1: Swap variables $x$ and $y$:
$$x = 4e^{2y - 3} + 5 \implies x - 5 = 4e^{2y - 3} \implies \frac{x - 5}{4} = e^{2y - 3}$$Step 2: Take natural logarithms of both sides:
$$\ln\left(\frac{x - 5}{4}\right) = 2y - 3$$ $$2y = 3 + \ln\left(\frac{x - 5}{4}\right) \implies y = \frac{3 + \ln\left(\frac{x - 5}{4}\right)}{2}$$ $$f^{-1}(x) = \frac{3 + \ln(x - 5) - \ln 4}{2}$$Step 3: Domain and range of $f^{-1}$:
Since $\text{range } f = (5, \infty)$, $\text{dom } f^{-1} = \{x \in \mathbb{R} \mid x > 5\}$.
Range of $f^{-1} = \mathbb{R}$.
Continuous Exponential Modeling: Half-Life & Doubling Time
- Continuous Growth / Decay Model: $$N(t) = N_0 e^{kt}$$ where $N_0 = N(0)$ is the initial quantity, $t$ is time, and $k$ is the continuous rate constant ($k > 0$ for growth, $k < 0$ for decay).
- Doubling Time ($T_d$): The time required for $N(t) = 2N_0$: $$2N_0 = N_0 e^{k T_d} \implies 2 = e^{k T_d} \implies T_d = \frac{\ln 2}{k}$$
- Half-Life ($t_{1/2}$): The time required for a decaying quantity to halve ($N(t) = 0.5N_0$): $$0.5N_0 = N_0 e^{k t_{1/2}} \implies 0.5 = e^{k t_{1/2}} \implies t_{1/2} = \frac{\ln 0.5}{k} = \frac{-\ln 2}{k} = \frac{\ln 2}{|k|}$$
(a) State the initial mass.
(b) Find the mass remaining after $25$ years.
(c) Find the exact half-life of the isotope in years.
Part (a): Initial mass
$$M(0) = 80e^0 = 80\text{ grams}$$Part (b): Mass after 25 years
$$M(25) = 80e^{-0.04(25)} = 80e^{-1} = \frac{80}{e} \approx 29.4\text{ grams}$$Part (c): Half-life
$$80e^{-0.04t} = 40 \implies e^{-0.04t} = 0.5$$ $$-0.04t = \ln(0.5) = -\ln 2 \implies t = \frac{\ln 2}{0.04} = \frac{\ln 2}{1/25} = 25\ln 2$$Answer: $t_{1/2} = 25\ln 2 \approx 17.3\text{ years}$.
Parameter Recovery & Half-Life Logarithmic Modeling
Initially, the mass is $160\text{ g}$. After $15$ days, the remaining mass is $100\text{ g}$.
(a) State the value of $M_0$. [1 mark]
(b) Find the exact value of the decay constant $k$. [3 marks]
(c) Determine the half-life of the isotope, giving your answer to $3$ significant figures. [3 marks]
Part (a): Initial mass $M_0$:
$$M(0) = 160 \implies M_0 = 160\text{ g} \quad [A1]$$Part (b): Find decay constant $k$:
$$M(15) = 160 e^{-15k} = 100 \implies e^{-15k} = \frac{100}{160} = \frac{5}{8} \quad [M1]$$Take natural logarithms of both sides:
$$-15k = \ln\left(\frac{5}{8} \right) \implies 15k = \ln\left(\frac{8}{5} \right) \quad [M1]$$ $$k = \frac{1}{15}\ln\left(\frac{8}{5} \right) \approx 0.031334\text{ day}^{-1} \quad [A1]$$Part (c): Half-life calculation (mass reduces to $\frac{1}{2}M_0 = 80\text{ g}$):
$$160 e^{-kt_{1/2}} = 80 \implies e^{-kt_{1/2}} = \frac{1}{2} \implies kt_{1/2} = \ln 2 \quad [M1]$$ $$t_{1/2} = \frac{\ln 2}{k} = \frac{\ln 2}{\frac{1}{15}\ln(8/5)} = \frac{15\ln 2}{\ln(1.6)} \quad [M1]$$ $$t_{1/2} = \frac{15(0.693147)}{0.470004} \approx 22.122 \implies 22.1\text{ days} \quad [A1]$$Answers: (a) $M_0 = 160\text{ g}$; (b) $k = \frac{1}{15}\ln(1.6)$; (c) $t_{1/2} \approx 22.1\text{ days}$
10 Progressive Practice Exercises
Tier 1 tests core exponential and logarithmic evaluations; Tier 2 introduces equation solving, asymptotes, and inverses; Tier 3 challenges quadratic substitution in $e^x$ and real-world cooling and bacterial models.
Tier 1 Core Fluency (Problems 1–3)
For the function $f(x) = 5 \cdot 3^x - 10$, state the equation of the horizontal asymptote and find the coordinates of the $y$-intercept and $x$-intercept.
View Full Worked Solution
Horizontal asymptote: $y = -10$.
$y$-intercept: $f(0) = 5(3^0) - 10 = 5(1) - 10 = -5 \implies (0, -5)$.
$x$-intercept: $5(3^x) - 10 = 0 \implies 3^x = 2 \implies x = \log_3 2 = \frac{\ln 2}{\ln 3} \approx 0.631 \implies (\log_3 2, 0)$.
Find the domain and the equation of the vertical asymptote of $f(x) = \ln(3x - 12)$.
View Full Worked Solution
Domain condition: $3x - 12 > 0 \implies 3x > 12 \implies x > 4$.
Domain: $\{x \in \mathbb{R} \mid x > 4\}$ or $(4, \infty)$.
Vertical asymptote: $x = 4$.
Find an expression for $f^{-1}(x)$ for $f(x) = e^{x + 2} - 3$.
View Full Worked Solution
Swap $x$ and $y$:
$$x = e^{y + 2} - 3 \implies x + 3 = e^{y + 2}$$ $$\ln(x + 3) = y + 2 \implies y = \ln(x + 3) - 2$$Answer: $f^{-1}(x) = \ln(x + 3) - 2$, with domain $x > -3$.
Tier 2 Exam Modeling (Problems 4–7)
Solve $4e^{3x - 1} = 28$. Give your answer in exact form and to 3 significant figures.
View Full Worked Solution
Step 1: Isolate the exponential:
$$e^{3x - 1} = \frac{28}{4} = 7$$Step 2: Take natural logarithms:
$$3x - 1 = \ln 7 \implies 3x = 1 + \ln 7 \implies x = \frac{1 + \ln 7}{3}$$Step 3: Approximate to 3 s.f.:
$$x \approx \frac{1 + 1.9459}{3} = \frac{2.9459}{3} \approx 0.982$$Answer: $x = \frac{1 + \ln 7}{3} \approx 0.982$.
Find the exact coordinates of the $x$-intercept of $f(x) = 2\ln(x + 3) - 4$.
View Full Worked Solution
Set $f(x) = 0$:
$$2\ln(x + 3) - 4 = 0 \implies 2\ln(x + 3) = 4 \implies \ln(x + 3) = 2$$ $$x + 3 = e^2 \implies x = e^2 - 3$$Answer: $(e^2 - 3, 0)$ (approximately $(4.39, 0)$).
An exponential curve $y = A e^{kx}$ passes through $(0, 12)$ and $(4, 3)$. Find the exact values of $A$ and $k$.
View Full Worked Solution
Step 1: Use point $(0, 12)$:
$$12 = A e^{k(0)} = A e^0 = A \implies A = 12$$Step 2: Use point $(4, 3)$:
$$3 = 12 e^{4k} \implies e^{4k} = \frac{3}{12} = \frac{1}{4} = 2^{-2}$$ $$4k = \ln\left(\frac{1}{4}\right) = -\ln 4 = -2\ln 2 \implies k = \frac{-2\ln 2}{4} = -\frac{1}{2}\ln 2$$Answer: $A = 12$ and $k = -\frac{1}{2}\ln 2 = \ln\left(\frac{1}{\sqrt{2}}\right) \approx -0.347$.
Solve the equation $e^{2x} - 5e^x + 6 = 0$ for $x$.
View Full Worked Solution
Let $u = e^x$ ($u > 0$):
$$u^2 - 5u + 6 = 0 \implies (u - 2)(u - 3) = 0$$ $$u = 2 \quad \text{or} \quad u = 3$$Substitute back $e^x = u$:
$$e^x = 2 \implies x = \ln 2$$ $$e^x = 3 \implies x = \ln 3$$Answer: $x = \ln 2$ or $x = \ln 3$.
Tier 3 Complex Applications & Synthesis (Problems 8–10)
Show that the line $y = x + 1$ is tangent to the curve $y = e^x$ at the point $(0, 1)$, and deduce that $e^x \ge x + 1$ for all real $x$.
View Full Worked Solution
Step 1: Verify common point at $x = 0$:
On line: $y = 0 + 1 = 1 \implies (0, 1)$.
On curve: $y = e^0 = 1 \implies (0, 1)$.
Step 2: Compare gradients:
Gradient of line is $m = 1$. Derivative of $e^x$ is $e^x$, which at $x = 0$ is $e^0 = 1$.
Since both the point $(0, 1)$ and the gradient $1$ match, the line is tangent to the curve at $(0, 1)$.
Step 3: Inequality deduction:
Because $y = e^x$ is strictly convex (concave up everywhere), the tangent line lies entirely below the curve for all $x \ne 0$, establishing $e^x \ge x + 1$ with equality only at $x = 0$. $\blacksquare$
A hot cup of coffee is left in a room with constant ambient temperature. Its temperature $T(t)$ in $^\circ\text{C}$ after $t$ minutes is modeled by:
$$T(t) = 20 + 70e^{-0.05t}, \quad t \ge 0$$
(a) State the initial temperature of the coffee.
(b) State the horizontal asymptote of $T(t)$ as $t \to \infty$ and interpret its physical meaning.
(c) Find the time taken for the coffee to cool to $45^\circ\text{C}$. Give your answer in exact form and to the nearest minute.
View Full Worked Solution
Part (a): Initial temperature
$$T(0) = 20 + 70e^0 = 20 + 70 = 90^\circ\text{C}$$Part (b): Horizontal Asymptote
$$\lim_{t \to \infty} (20 + 70e^{-0.05t}) = 20 + 0 = 20 \implies T = 20$$Physical meaning: Over a long time, the coffee cools down until it reaches the ambient room temperature of $20^\circ\text{C}$.
Part (c): Time to reach $45^\circ\text{C}$
$$20 + 70e^{-0.05t} = 45 \implies 70e^{-0.05t} = 25 \implies e^{-0.05t} = \frac{25}{70} = \frac{5}{14}$$ $$-0.05t = \ln\left(\frac{5}{14}\right) \implies t = \frac{\ln(14/5)}{0.05} = 20\ln(2.8) \approx 20(1.0296) \approx 20.6\text{ minutes}$$Answer: (a) $90^\circ\text{C}$; (b) $T = 20^\circ\text{C}$; (c) $20\ln(2.8) \approx 21\text{ minutes}$.
A bacterial culture has an initial population of $200$ cells and grows to $1600$ cells after $6\text{ hours}$ according to the model $P(t) = P_0 e^{kt}$.
(a) Find the exact value of the growth constant $k$.
(b) Determine the exact doubling time of the bacterial population.
(c) Find the time required for the population to reach $10{,}000$ cells, to 3 significant figures.
View Full Worked Solution
Part (a): Growth constant $k$
$$P(0) = P_0 = 200$$ $$1600 = 200e^{6k} \implies e^{6k} = 8 = 2^3 \implies 6k = \ln(2^3) = 3\ln 2 \implies k = \frac{\ln 2}{2} = 0.5\ln 2 \approx 0.347$$Part (b): Doubling time
$$T_d = \frac{\ln 2}{k} = \frac{\ln 2}{\frac{\ln 2}{2}} = 2\text{ hours}$$Part (c): Time to reach 10,000 cells
$$200e^{kt} = 10000 \implies e^{kt} = 50 \implies kt = \ln 50$$ $$t = \frac{\ln 50}{k} = \frac{\ln 50}{0.5\ln 2} = \frac{2\ln 50}{\ln 2} \approx \frac{2(3.9120)}{0.6931} \approx 11.3\text{ hours}$$Answer: (a) $k = 0.5\ln 2$; (b) $2\text{ hours}$; (c) $11.3\text{ hours}$.
Logarithmic Laws & Identities
- Product Law: $$\log_a(xy) = \log_a x + \log_a y$$
- Quotient Law: $$\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y$$
- Power Law: $$\log_a(x^k) = k \log_a x$$
- Change of Base: $$\log_a x = \frac{\log_b x}{\log_b a} = \frac{\ln x}{\ln a}$$
Exponentials, Inverses & Asymptotes
- Natural Log / Exponential Duality: $$y = e^x \iff x = \ln y \quad (y > 0)$$ $$e^{\ln x} = x, \quad \ln(e^x) = x$$
- Horizontal Asymptote of $y = A e^{bx} + c$: $$y = c$$
- Vertical Asymptote of $y = A \ln(bx - h) + c$: $$x = \frac{h}{b}$$
- Continuous Doubling Time & Half-Life: $$T_d = \frac{\ln 2}{k}, \qquad t_{1/2} = \frac{\ln 2}{|k|}$$
Critical IB Examination Pitfalls
- Non-Positive Logarithm Arguments: Always check solutions against the original domain. $\ln(x)$ is only defined for $x > 0$. If an algebraic step yields $e^x = -3$, reject it immediately.
- Confusing $\ln(A + B)$ with $\ln A + \ln B$: The logarithm of a sum CANNOT be split. $\ln(x + y) \ne \ln x + \ln y$.
- Neglecting Asymptote Shifts: For $f(x) = 2e^{3x} - 5$, the horizontal asymptote is $y = -5$, NOT $y = 0$. The range is $y > -5$.
- Misidentifying Base $e$: $e \approx 2.71828$ is an irrational mathematical constant, not a variable. $\ln x \equiv \log_e x$.
IB AA SL Topic Check: Exponential & Logarithmic Functions
Section A has diagnostic questions with feedback. Section B has Paper 1 and Paper 2 exam-style problems with markschemes.
Section A: Diagnostic Concept Check
Score: 0 / 4Question 1: What is the equation of the horizontal asymptote of $y = 3e^{-2x} + 4$?
Question 2: What is the domain of the function $f(x) = \ln(8 - 2x)$?
Question 3: What is the exact solution to the equation $e^{2x} = 5$?
Question 4: What are the coordinates of the $y$-intercept of $y = 2^{x + 3} - 5$?
Section B: Exam-Style Free Response Questions
Question 5: Transformed Exponential Curve
Consider the function $f(x) = 3e^{2x} - 6$.
(a) Write down the equation of the horizontal asymptote of the graph of $f$. [1 mark]
(b) Find the exact value of the $x$-intercept. [2 marks]
(c) Find an expression for $f^{-1}(x)$ and state its domain. [3 marks]
View Official Markscheme & Model Solution
Part (a) [1 Mark]:
Horizontal asymptote: $y = -6$ (A1)
Part (b) [2 Marks]:
$$3e^{2x} - 6 = 0 \implies 3e^{2x} = 6 \implies e^{2x} = 2 \quad \text{\bf (M1)}$$ $$2x = \ln 2 \implies x = \frac{1}{2}\ln 2 \quad (\text{or } \ln\sqrt{2}) \quad \text{\bf (A1)}$$Part (c) [3 Marks]:
$$x = 3e^{2y} - 6 \implies x + 6 = 3e^{2y} \implies e^{2y} = \frac{x + 6}{3} \quad \text{\bf (M1)}$$ $$2y = \ln\left(\frac{x + 6}{3}\right) \implies f^{-1}(x) = \frac{1}{2}\ln\left(\frac{x + 6}{3}\right) \quad \text{\bf (A1)}$$Domain of $f^{-1}$: $\{x \in \mathbb{R} \mid x > -6\}$ (A1)
Question 6: Population Modeling & Doubling Time
A population of endangered mammals in a nature reserve is modeled by $P(t) = 500e^{0.08t}$, where $t$ is the time in years after conservation efforts began.
(a) State the population at the start of the conservation project. [1 mark]
(b) Find the population after $10$ years, to the nearest integer. [2 marks]
(c) Find the exact time required for the population to double in size, and calculate this value to 3 significant figures. [4 marks]
View Official Markscheme & Model Solution
Part (a) [1 Mark]:
$$P(0) = 500 \quad \text{\bf (A1)}$$Part (b) [2 Marks]:
$$P(10) = 500e^{0.08(10)} = 500e^{0.8} \quad \text{\bf (M1)}$$ $$P(10) \approx 500(2.22554) \approx 1113\text{ mammals} \quad \text{\bf (A1)}$$Part (c) [4 Marks]:
Doubling condition: $P(t) = 1000$ (M1)
$$500e^{0.08t} = 1000 \implies e^{0.08t} = 2 \quad \text{\bf (A1)}$$ $$0.08t = \ln 2 \implies t = \frac{\ln 2}{0.08} = \frac{\ln 2}{2/25} = \frac{25\ln 2}{2} \quad \text{\bf (A1)}$$ $$t \approx 8.66\text{ years} \quad \text{\bf (A1)}$$Topic 2 Curriculum
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About this guide
Independent SJMaths resource; not an official IB publication. Follows the AA syllabus first assessed in 2021 (final assessment: November 2028). Use the separate 2029 guide for the new course.
Check the IB Mathematics programme page and the official AA subject guide for the authoritative syllabus and examination-session requirements.