Comprehensive assessment covering all 15 lessons: exponential and logarithmic functions, log laws, change of base, differentiation of $e^x$, $a^x$, $\ln x$, and $\log_a x$, growth and decay, and optimisation.
Assessment
Select the best answer for each question.
The value of $\log_2(32)$ is:
$\frac{d}{dx}(e^{5x})$ equals:
The domain of $y = \ln(x - 3)$ is:
$\frac{d}{dx}(2^{3x})$ equals:
Simplify $\log_3(27) - \log_3(9)$:
The solution to $\log_5(x - 2) = 2$ is:
$\frac{d}{dx}(\log_2 x)$ equals:
A population grows from 200 to 800 in 6 hours. The doubling time is:
The maximum value of $f(x) = 2xe^{-x}$ for $x \geq 0$ occurs at:
Using change of base, $\log_4(20)$ equals:
Short Answer
Differentiate $f(x) = x^2 e^{3x}$, showing all working and factorising your answer.
Solve $\log_2(x) + \log_2(x - 2) = 3$, showing all working and checking for extraneous solutions.
A drug has concentration $C(t) = 4te^{-0.2t}$ mg/L. Find when the maximum concentration occurs and calculate this maximum to 2 decimal places.
Use change of base to evaluate $\log_7(300)$ to 2 decimal places. Show your working.
Find the equation of the tangent to $y = \ln x$ at $x = e^2$. Give your answer in the form $y = mx + c$.
Q1: C — $2^5 = 32$, so $\log_2(32) = 5$.
Q2: B — $\frac{d}{dx}(e^{5x}) = 5e^{5x}$ by chain rule.
Q3: B — Domain requires $x - 3 > 0$, so $x > 3$.
Q4: C — $\frac{d}{dx}(2^{3x}) = 2^{3x} \cdot \ln 2 \cdot 3 = 3\ln(2) \cdot 2^{3x}$.
Q5: C — $\log_3(27) = 3$, $\log_3(9) = 2$, difference = 1.
Q6: C — $x - 2 = 5^2 = 25$, so $x = 27$.
Q7: C — $\frac{d}{dx}(\log_2 x) = \frac{1}{x \ln 2}$.
Q8: C — $800 = 200 \cdot 2^2$, so 2 doublings in 6 hours = 1 doubling in 3 hours.
Q9: B — $f'(x) = 2e^{-x}(1-x) = 0 \Rightarrow x = 1$.
Q10: B — Change of base: $\frac{\ln 20}{\ln 4}$.
Q11 (3 marks): $u = x^2$, $v = e^{3x}$ [0.5]. $u' = 2x$, $v' = 3e^{3x}$ [0.5]. $f'(x) = 2xe^{3x} + x^2 \cdot 3e^{3x}$ [0.5] $= xe^{3x}(2 + 3x)$ [1.5].
Q12 (3 marks): $\log_2(x(x-2)) = 3$ [0.5]. $x(x-2) = 8$ [0.5]. $x^2 - 2x - 8 = 0$ [0.5]. $(x-4)(x+2) = 0$ [0.5]. $x = 4$ or $x = -2$. Check: $x = 4$: $\log_2(4) + \log_2(2) = 2 + 1 = 3$ ✓ [0.25]. $x = -2$: $\log_2(-2)$ undefined ✗ [0.25]. Valid: $x = 4$ [0.5].
Q13 (3 marks): $C'(t) = 4e^{-0.2t} + 4t(-0.2)e^{-0.2t}$ [0.5] $= 4e^{-0.2t}(1 - 0.2t)$ [0.5]. $C'(t) = 0$ when $1 - 0.2t = 0$, so $t = 5$ hours [0.5]. $C(5) = 4(5)e^{-1} = 20e^{-1} = \frac{20}{e} \approx 7.36$ mg/L [1.5].
Q14 (3 marks): $\log_7(300) = \frac{\ln 300}{\ln 7}$ [1] $= \frac{5.704}{1.946}$ [0.5] $\approx 2.93$ [1.5].
Q15 (3 marks): At $x = e^2$: $y = \ln(e^2) = 2$ [0.5]. $\frac{dy}{dx} = \frac{1}{x}$ [0.5]. At $x = e^2$: $m = \frac{1}{e^2}$ [0.5]. Equation: $y - 2 = \frac{1}{e^2}(x - e^2)$ [0.5] $\Rightarrow y = \frac{x}{e^2} - 1 + 2 = \frac{x}{e^2} + 1$ [1].