Covers Lessons 6–10: logarithmic graphs, laws of logarithms, change of base, logarithmic equations, and differentiating $e^x$ and $a^x$. This checkpoint bridges the algebra of logarithms with the calculus of exponentials.
Assessment
Select the best answer for each question.
The vertical asymptote of $y = \log_2(x - 3) + 1$ is:
$\log_3(27) + \log_3(9)$ equals:
Using change of base, $\log_5(20)$ equals:
The solution to $\log_2(x - 1) = 3$ is:
$\frac{d}{dx}(e^{4x})$ equals:
$\frac{d}{dx}(3^x)$ equals:
For which value of $x$ does $\ln(x) + \ln(x - 3) = \ln(10)$ have a valid solution?
The gradient of $y = e^x$ at $x = 0$ is:
Short Answer
Sketch $y = \ln(x - 2) + 1$, labelling the vertical asymptote, $x$-intercept, and the point where $x = 3$.
Simplify $\log_2(32) + 2\log_2(4) - \log_2(8)$, showing each step clearly.
Use change of base to evaluate $\log_7(200)$ to 2 decimal places. Show your working.
Q1: C — The argument $x - 3 = 0$ when $x = 3$, so the vertical asymptote is $x = 3$.
Q2: C — $\log_3(27) = 3$ and $\log_3(9) = 2$, so sum = 5. Or: $\log_3(27 \times 9) = \log_3(243) = \log_3(3^5) = 5$.
Q3: B — Change of base: $\log_5(20) = \frac{\ln(20)}{\ln(5)}$.
Q4: B — $x - 1 = 2^3 = 8$, so $x = 9$.
Q5: C — $\frac{d}{dx}(e^{4x}) = 4e^{4x}$ by the chain rule.
Q6: C — $\frac{d}{dx}(3^x) = 3^x \ln(3)$.
Q7: C — $\ln(x) + \ln(x-3) = \ln(10) \Rightarrow x(x-3) = 10 \Rightarrow x^2 - 3x - 10 = 0 \Rightarrow (x-5)(x+2) = 0$. Need $x > 3$, so $x = 5$.
Q8: B — $\frac{d}{dx}(e^x) = e^x$, so at $x = 0$, gradient = $e^0 = 1$.
Q9 (3 marks): Asymptote: $x = 2$ [1]. $x$-intercept: $\ln(x-2) = -1 \Rightarrow x - 2 = e^{-1} \Rightarrow x = 2 + \frac{1}{e} \approx 2.37$, so $(2 + e^{-1}, 0)$ [1]. At $x = 3$: $y = \ln(1) + 1 = 0 + 1 = 1$, so $(3, 1)$ [1].
Q10 (3 marks): $\log_2(32) = 5$ [0.5], $\log_2(4) = 2$ [0.5], $\log_2(8) = 3$ [0.5]. Expression = $5 + 2(2) - 3 = 5 + 4 - 3 = 6$ [1.5].
Q11 (3 marks): $\log_7(200) = \frac{\ln(200)}{\ln(7)}$ [1] $= \frac{5.298}{1.946}$ [0.5] $\approx 2.72$ [1.5].