MYP Inquiry Framework
Key Concept Relationships
Related Concepts Change, Models, Representation
Global Context Globalization and sustainability (Population growth, carbon emissions & economic compounding)
Statement of Inquiry Modeling multiplicative change through exponential relationships reveals long-term boundaries and informs sustainable decision-making.
4.1

Anatomy of the Exponential Function $f(x) = a \cdot b^x + c$

  • Base Constraints: In $f(x) = a \cdot b^x + c$, the base $b$ must be strictly positive and not equal to one ($b > 0, b \ne 1$). If $b = 1$, the function degenerates into a constant horizontal line $f(x) = a + c$.
  • Horizontal Asymptote: As $x \to -\infty$ (for $b > 1$), $b^x \to 0$. The graph approaches but never crosses the horizontal line: $$\mathbf{y = c}$$
  • Initial Value / $y$-Intercept: Setting $x = 0$ gives $f(0) = a \cdot b^0 + c = a(1) + c = \mathbf{a + c}$. The $y$-intercept is $(0, a + c)$.
  • Domain and Range:
    • Domain: $x \in \mathbb{R}$ ($-\infty, \infty$).
    • Range (if $a > 0$): $y \in (c, \infty)$ or $y > c$.
    • Range (if $a < 0$, reflected across horizontal): $y \in (-\infty, c)$ or $y < c$.
x y Asymptote: y = 0 Growth: y = b^x (b > 1) Decay: y = b^x (0 < b < 1) (0, 1)
4.2

Exponential Growth vs. Exponential Decay

  • Exponential Growth ($b > 1$): The rate of change is proportional to the current amount: the larger the quantity grows, the faster it increases. As $x \to \infty$, $f(x) \to \infty$. Typical base representations: $b = 1 + r$, where $r$ is the fractional growth rate per period.
  • Exponential Decay ($0 < b < 1$): Can also be written as $b = c^{-1}$ or $f(x) = a \cdot c^{-x}$ with $c > 1$. The quantity diminishes rapidly at first, then flattens asymptotically towards $y = c$. Typical base representation: $b = 1 - r$, where $r$ is the depreciation rate.
  • Comparison with Linear and Quadratic Models:
    • Linear ($y = mx + c$): Constant difference (arithmetic progression).
    • Quadratic ($y = ax^2 + bx + c$): Constant second difference (accelerated polynomial growth).
    • Exponential ($y = a \cdot b^x$): Constant ratio (geometric progression); eventually outgrows any polynomial for sufficiently large $x$.
4.3

Real-World Models: Finance & Science

  • Compound Interest Formula: $$A = P\left(1 + \frac{r}{k}\right)^{kt}$$ Where $P$ is principal, $r$ is annual nominal interest rate in decimal form, $k$ is number of compounding periods per year ($k=1$ annually, $k=4$ quarterly, $k=12$ monthly), and $t$ is time in years.
  • Depreciation / Reducing Balance: $$V = V_0(1 - r)^t$$ Where $V_0$ is the purchase price and $r$ is the annual rate of depreciation.
  • Radioactive Half-Life Model: $$N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}$$ Where $N_0$ is the initial sample mass, $h$ is the half-life duration, and $t$ is elapsed time in the same units.
4.4

Solving Exponential Equations

  • Method 1: Equating Like Bases: If $b^u = b^v$ and $b > 0, b \ne 1$, then by one-to-one property: $$\mathbf{u = v}$$ Example: $8^{x - 1} = 4^{x + 2} \implies (2^3)^{x-1} = (2^2)^{x+2} \implies 2^{3x-3} = 2^{2x+4} \implies 3x - 3 = 2x + 4 \implies x = 7$.
  • Method 2: Using Logarithms for Unlike Bases: If $a^x = b$ where bases cannot be easily matched, take the common or natural logarithm of both sides: $$\log(a^x) = \log(b) \implies x \log(a) = \log(b) \implies \mathbf{x = \frac{\log b}{\log a}}$$
  • Method 3: Quadratic Substitution ($u = b^x$): Equations of the form $a \cdot b^{2x} + c \cdot b^x + d = 0$. Substitute $u = b^x$ to obtain $a u^2 + c u + d = 0$, solve for $u > 0$, then find $x$.

Worked Examples (Criterion A & C)

Example 1 Solving an Exponential Equation via Base Reduction

Problem: Solve for $x$: $9^{2x - 1} = \left(\dfrac{1}{27}\right)^{x + 2}$.

Step 1: Express all terms with a common base of $3$

$9 = 3^2$ and $\frac{1}{27} = 27^{-1} = (3^3)^{-1} = 3^{-3}$.

Step 2: Apply index law of powers $(b^m)^n = b^{mn}$ $$(3^2)^{2x - 1} = (3^{-3})^{x + 2} \implies 3^{2(2x - 1)} = 3^{-3(x + 2)}$$ $$3^{4x - 2} = 3^{-3x - 6}$$
Step 3: Equate exponents and solve the linear equation $$4x - 2 = -3x - 6 \implies 4x + 3x = -6 + 2 \implies 7x = -4$$ $$\mathbf{x = -\frac{4}{7}}$$ Criterion A: Complete index law application with exact fractional solution
Example 2 Financial Compound Interest Investment Modeling

Problem: Elena invests $\$5000$ into an account earning $6\%$ annual interest compounded monthly.
(a) Write an exponential function $A(t)$ giving the balance after $t$ years.
(b) Calculate the account balance after $5$ years.
(c) Use logarithms to find how many full years it will take for the investment to double in value.

Step 1: Formulate the model function

Principal $P = 5000$, $r = 0.06$, compounding frequency $k = 12$ months per year.

$$A(t) = 5000\left(1 + \frac{0.06}{12}\right)^{12t} = \mathbf{5000(1.005)^{12t}}$$
Step 2: Evaluate balance after $5$ years ($t = 5$) $$A(5) = 5000(1.005)^{60} \approx 5000 \times 1.34885 = \mathbf{\$6744.25}$$
Step 3: Solve for doubling time ($A(t) = 10000$) $$5000(1.005)^{12t} = 10000 \implies (1.005)^{12t} = 2$$

Take logarithms of both sides:

$$\log\left((1.005)^{12t}\right) = \log(2) \implies 12t \cdot \log(1.005) = \log(2)$$ $$t = \frac{\log(2)}{12 \cdot \log(1.005)} \approx \frac{0.30103}{12 \times 0.002166} \approx \frac{0.30103}{0.02599} \approx 11.58\text{ years}$$

Therefore, it will take $\mathbf{12\text{ full years}}$ for the investment to double.

Criterion C & D: Authentic real-world contextualization with correct financial rounding

MYP 4–5 Practice Exercises: Exponential Functions

Tier 1 (Core skills), Tier 2 (Reasoning), and Tier 3 (Extension and Criterion D modelling).

Question 1 Tier 1: Standard 3 Marks

For the function $f(x) = 3 \cdot 2^x - 5$, find:
(a) The equation of the horizontal asymptote.
(b) The coordinates of the $y$-intercept.

View Full Method Markscheme
[M1] In $f(x) = a \cdot b^x + c$, as $x \to -\infty$, $2^x \to 0$.
[A1] Part (a): The horizontal asymptote is $\mathbf{y = -5}$.
[M1] Part (b): Substitute $x = 0$: $f(0) = 3(2^0) - 5 = 3(1) - 5 = -2$.
[A1] $y$-intercept is $\mathbf{(0, -2)}$.
Question 2 Tier 1: Standard 3 Marks

Solve the exponential equation by equating bases: $5^{2x + 1} = 125$.

View Full Method Markscheme
[M1] Write $125$ as a power of $5$: $125 = 5^3$.
[M1] Equate exponents: $2x + 1 = 3$.
[A1] Solve for $x$: $2x = 2 \implies \mathbf{x = 1}$.
Question 3 Tier 1: Standard 3 Marks

A new car costs $\$28000$ and depreciates by $15\%$ per annum. Calculate its estimated value after $4$ years.

View Full Method Markscheme
[M1] Decay multiplier is $1 - 0.15 = 0.85$.
[M1] Set up model: $V(4) = 28000(0.85)^4$.
[A1] Compute $(0.85)^4 \approx 0.522006$. Value: $28000 \times 0.522006 \approx \mathbf{\$14616.18}$ (or $\$14600$ to 3 s.f.).
Question 4 Tier 2: Intermediate 4 Marks

Solve for $x$: $4^{x - 1} \cdot 8^{2x} = 16^{x + 3}$.

View Full Method Markscheme
[M1] Convert all bases to $2$: $4 = 2^2, \; 8 = 2^3, \; 16 = 2^4$.
[M1] Substitute: $(2^2)^{x-1} \cdot (2^3)^{2x} = (2^4)^{x+3} \implies 2^{2x-2} \cdot 2^{6x} = 2^{4x+12}$.
[M1] Combine left side using index multiplication law: $2^{(2x-2) + 6x} = 2^{8x-2}$.
[A1] Equate exponents: $8x - 2 = 4x + 12 \implies 4x = 14 \implies \mathbf{x = \frac{7}{2} = 3.5}$.
Question 5 Tier 2: Intermediate 4 Marks

An exponential curve $y = a \cdot b^x$ passes through the points $(1, 6)$ and $(3, 24)$. Find the values of constants $a$ and $b$ (with $b > 0$).

View Full Method Markscheme
[M1] Set up simultaneous equations: $$(1): \; 6 = a \cdot b^1, \qquad (2): \; 24 = a \cdot b^3$$
[M1] Divide equation (2) by equation (1): $$\frac{24}{6} = \frac{a \cdot b^3}{a \cdot b} \implies 4 = b^2$$
[A1] Since base $b > 0$, $\mathbf{b = 2}$.
[A1] Substitute into (1): $6 = a(2) \implies \mathbf{a = 3}$. Curve is $y = 3 \cdot 2^x$.
Question 6 Tier 2: Intermediate 4 Marks

Solve for $x$ giving your answer to 3 significant figures: $7^x = 45$.

View Full Method Markscheme
[M1] Take the base-10 logarithm of both sides: $\log_{10}(7^x) = \log_{10}(45)$.
[M1] Apply power rule: $x \log_{10}(7) = \log_{10}(45)$.
[M1] Isolate $x$: $x = \frac{\log_{10}(45)}{\log_{10}(7)}$.
[A1] Evaluate using calculator: $x \approx \frac{1.65321}{0.84510} \approx \mathbf{1.96}$.
Question 7 Tier 2: Intermediate 4 Marks

Solve the quadratic-type exponential equation: $2^{2x} - 10 \cdot 2^x + 16 = 0$.

View Full Method Markscheme
[M1] Let $u = 2^x$. Then $2^{2x} = (2^x)^2 = u^2$.
[M1] Form quadratic: $u^2 - 10u + 16 = 0 \implies (u - 2)(u - 8) = 0$.
[M1] Solutions for $u$: $u = 2$ or $u = 8$.
[A1] Back-substitute:
  • $2^x = 2^1 \implies \mathbf{x = 1}$
  • $2^x = 8 = 2^3 \implies \mathbf{x = 3}$
Question 8 Tier 3: Extended 5 Marks

Criterion D Radioactive Decay Investigation: A sample of Iodine-131 has an initial mass of $160\text{ mg}$ and decays with a half-life of $8\text{ days}$.
(a) Write a formula for remaining mass $M(t)$ at time $t$ days.
(b) Calculate remaining mass after $24\text{ days}$.
(c) Determine algebraically after how many days the mass will drop below $5\text{ mg}$.

View Full Method Markscheme
[M1] Part (a): Half-life model: $\mathbf{M(t) = 160 \left(\frac{1}{2}\right)^{t / 8}}$.
[M1] Part (b): At $t = 24$, $t / 8 = 3$ half-lives: $M(24) = 160 \left(\frac{1}{2}\right)^3 = 160 \times \frac{1}{8} = \mathbf{20\text{ mg}}$.
[M1] Part (c): Set $M(t) = 5$: $160 \left(\frac{1}{2}\right)^{t / 8} = 5 \implies \left(\frac{1}{2}\right)^{t / 8} = \frac{5}{160} = \frac{1}{32}$.
[M1] Recognise $32 = 2^5$, so $\left(\frac{1}{2}\right)^{t / 8} = \left(\frac{1}{2}\right)^5$.
[A1] Equate exponents: $\frac{t}{8} = 5 \implies \mathbf{t = 40\text{ days}}$. Thus mass drops below $5\text{ mg}$ after strictly more than $40$ days.
Question 9 Tier 3: Extended 5 Marks

Compare the long-term behaviour of three models for $x \ge 0$: $$f(x) = 200x + 500, \quad g(x) = 2x^2 + 50, \quad h(x) = 5 \cdot (1.5)^x$$ (a) Determine which function has the largest value at $x = 2$.
(b) Determine which function has the largest value at $x = 10$.
(c) Explain why $h(x)$ will eventually exceed both $f(x)$ and $g(x)$ for all sufficiently large $x$.

View Full Method Markscheme
[M1] Evaluate at $x = 2$: $f(2) = 900$, $g(2) = 2(4)+50 = 58$, $h(2) = 5(1.5)^2 = 11.25$. Largest is $\mathbf{f(x)}$.
[M1] Evaluate at $x = 10$: $f(10) = 2500$, $g(10) = 2(100)+50 = 250$, $h(10) = 5(1.5)^{10} \approx 5(57.665) = 288.3$. Largest is still $\mathbf{f(x)}$.
[M1] Evaluate at e.g. $x = 20$: $f(20) = 4500$, $g(20) = 850$, $h(20) = 5(1.5)^{20} \approx 5(3325.3) = 16626.5$. Here $h(x)$ dominates.
[R1] Part (c): In linear models, the rate of change is constant. In quadratics, rate of change is linear. In an exponential function with $b > 1$, the rate of change is proportional to the function itself (geometric acceleration). Therefore, the exponential curve will always dominate any polynomial function as $x \to \infty$.
Question 10 Tier 3: Extended 5 Marks

Solve for all real values of $x$: $3^{x + 1} + 3^{2 - x} = 28$.

View Full Method Markscheme
[M1] Decompose exponents using index laws: $3^{x+1} = 3 \cdot 3^x$, and $3^{2-x} = \frac{3^2}{3^x} = \frac{9}{3^x}$.
[M1] Let $u = 3^x$. Substitute into equation: $3u + \frac{9}{u} = 28$.
[M1] Multiply through by $u$: $3u^2 - 28u + 9 = 0$.
[M1] Factorise quadratic: $ac = 27$. Factors $-27$ and $-1$: $(3u - 1)(u - 9) = 0 \implies u = \frac{1}{3}$ or $u = 9$.
[A1] Back-substitute $u = 3^x$:
  • $3^x = \frac{1}{3} = 3^{-1} \implies \mathbf{x = -1}$
  • $3^x = 9 = 3^2 \implies \mathbf{x = 2}$

Exponential Functions Formulas

  • Standard Form: $f(x) = a \cdot b^x + c \quad (b > 0, \, b \ne 1)$
  • Horizontal Asymptote: $y = c$
  • $y$-Intercept: $(0, a + c)$
  • Compound Interest: $A = P\left(1 + \dfrac{r}{k}\right)^{kt}$
  • Continuous Growth/Decay: $N(t) = N_0 e^{kt}$
  • Half-Life Model: $N(t) = N_0\left(\dfrac{1}{2}\right)^{t / h}$
  • Logarithm Change of Base: $b^x = A \iff x = \dfrac{\log A}{\log b}$

Top Common MYP Pitfalls

1. Violating Order of Operations

In $3 \cdot 2^x$, multiplying $3 \times 2$ to write $6^x$. Exponentiation has precedence over multiplication! $3 \cdot 2^x \ne 6^x$.

2. Mistaking Decay Base

In depreciation of $12\%$, using base $b = 0.12$. The decay multiplier is $1 - 0.12 = 0.88$!

3. Claiming Exponential Curves Have $x$-Intercepts Everywhere

The standard curve $y = 2^x$ never touches or crosses the $x$-axis because $2^x > 0$ for all real $x$.

4. Compounding Period Mistakes

In monthly compounding ($k = 12$), dividing the rate by $12$ but forgetting to multiply years $t$ by $12$ in the exponent.

Interactive Mini Test: Exponential Functions

Answer the questions to see your score and feedback.

Question 1 of 5 Horizontal Asymptote

What is the equation of the horizontal asymptote for $f(x) = 4(0.5)^x + 7$?

Question 2 of 5 Solving Exponential Equation

Solve for $x$: $27^{x - 1} = 9^{x + 2}$.

Question 3 of 5 Decay Multiplier

A population of bacteria decreases by $8\%$ every hour. What is the hourly decay factor (base $b$)?

Question 4 of 5 Compound Interest Compounding Frequency

If $\$1000$ is invested at $4\%$ per year compounded quarterly for $3$ years, which expression calculates the final balance?

Question 5 of 5 Range of Exponential Function

What is the range of the function $g(x) = -2(3)^x + 10$?