This checkpoint assesses your understanding of exponential functions, their graphs, the number $e$, exponential modelling, and logarithmic functions (Lessons 1–5). It covers domain, range, transformations, solving equations, and real-world applications.
1. The range of $y = 3^x$ is: 1 MARK
2. The horizontal asymptote of $y = 2^x - 3$ is: 1 MARK
3. The value of $e$ is approximately: 1 MARK
4. Solve $4^x = 64$. 1 MARK
5. The domain of $y = \log_5(x)$ is: 1 MARK
6. $\log_2(32)$ equals: 1 MARK
7. A population is modelled by $P(t) = 200e^{0.08t}$. The population after 5 years is approximately: 1 MARK
8. The graph of $y = 3^{-x}$ is obtained from $y = 3^x$ by: 1 MARK
9. (a) Sketch $y = 2^{x+1} - 3$, labelling the horizontal asymptote, $y$-intercept, and $x$-intercept (if it exists). (b) Describe the sequence of transformations from $y = 2^x$ to $y = 2^{x+1} - 3$. (c) State the domain and range of $y = 2^{x+1} - 3$. 4 MARKS
10. A sample of radioactive material decays from 100g to 25g in 20 days. (a) Find the decay constant $k$ assuming $A = A_0 e^{kt}$. (b) Find the half-life. (c) How much remains after 30 days? 4 MARKS
11. The pH scale is logarithmic: $\text{pH} = -\log_{10}[\text{H}^+]$. A student claims that since the difference between pH 2 and pH 3 is just 1 unit, the acidity difference is small. Evaluate this claim, using specific calculations to show the actual difference in hydrogen ion concentration. 4 MARKS
(a) Asymptote: $y = -3$ [0.5]. $y$-intercept: when $x = 0$, $y = 2^1 - 3 = -1$, so $(0, -1)$ [0.5]. $x$-intercept: $2^{x+1} - 3 = 0 \Rightarrow 2^{x+1} = 3 \Rightarrow x+1 = \log_2 3 \Rightarrow x = \log_2 3 - 1 \approx 0.585$ [0.5]. Sketch shows growth curve approaching $y = -3$ from above, passing through $(0, -1)$ and $(0.585, 0)$ [0.5].
(b) Start with $y = 2^x$. Shift left by 1 unit: $y = 2^{x+1}$ [0.5]. Then shift down by 3 units: $y = 2^{x+1} - 3$ [0.5].
(c) Domain: all real numbers ($x \in \mathbb{R}$) [0.5]. Range: $y > -3$ [0.5].
(a) $25 = 100e^{20k} \Rightarrow e^{20k} = 0.25 \Rightarrow 20k = \ln(0.25) = -\ln 4 \approx -1.386$ [1]. So $k = \frac{-1.386}{20} \approx -0.0693$ per day [0.5].
(b) $t_{1/2} = \frac{\ln 2}{|k|} = \frac{0.693}{0.0693} = 10$ days [1]. (Alternatively: 100g → 50g → 25g takes 20 days, so half-life = 10 days.)
(c) $A(30) = 100e^{-0.0693 \times 30} = 100e^{-2.079} \approx 100 \times 0.125 = 12.5$ g [1]. (Or: 30 days = 3 half-lives, so $100 \times (\frac{1}{2})^3 = 12.5$ g.)
The student's claim is fundamentally incorrect [1]. The pH scale is logarithmic, not linear. At pH 2: $[\text{H}^+] = 10^{-2} = 0.01$ mol/L. At pH 3: $[\text{H}^+] = 10^{-3} = 0.001$ mol/L [1]. The ratio is $\frac{10^{-2}}{10^{-3}} = 10^1 = 10$ [1]. This means a solution with pH 2 is actually 10 times more acidic than one with pH 3 — a substantial difference, not a small one. The logarithmic nature of the pH scale compresses a huge range of concentrations (spanning 14 orders of magnitude from $10^0$ to $10^{-14}$) into a simple 0–14 scale, making small pH differences represent large actual differences [1].