Level 3 — ProductionExponentials & Logarithms

Exponentials & Logarithms

⏱ 45 minutes60 marksprintable — key stays hidden on paper

Level: 3 (Production — from-scratch derivations, explain-out-loud, code-from-memory) Time limit: 45 minutes Total marks: 60

Instructions: Show every step. Where asked to "prove" or "derive", you may not simply quote the result — you must build it. Where asked to "explain out loud", write your reasoning as full sentences.

Question 1 — Derive the laws of logarithms (12 marks)

Starting only from the definition of a logarithm as the inverse of an exponential function (i.e. $y=\log_a x \iff a^y = x$ , with $a>0,\ a\neq 1$ ):

(a) Prove the product rule: $\log_a(MN) = \log_a M + \log_a N$ . (4) (b) Prove the power rule: $\log_a(M^k) = k\log_a M$ . (4) (c) Prove the change-of-base formula: $\log_a x = \dfrac{\log_b x}{\log_b a}$ . (4)

Question 2 — The number $e$ (8 marks)

(a) State the limit definition of $e$ and explain out loud, in the context of continuously compounded growth, why $\left(1+\tfrac1n\right)^n$ approaches a fixed number as $n\to\infty$ . (4) (b) A bank offers a nominal annual rate of $100\%$ . Compute the value of $\left(1+\tfrac1n\right)^n$ for $n=1,2,4,12$ and comment on the trend relative to $e\approx 2.71828$ . (4)

Question 3 — Solving exponential & logarithmic equations (12 marks)

(a) Solve $5^{2x-1} = 3^{x+2}$ , giving $x$ exactly in terms of natural logarithms and then to 3 s.f. (5) (b) Solve $\log_2(x) + \log_2(x-2) = 3$ . State and justify any rejected roots. (4) (c) Solve $e^{2x} - 5e^{x} + 6 = 0$ . (3)

Question 4 — Growth & decay models (12 marks)

A radioactive isotope decays according to $N(t) = N_0 e^{-kt}$ , where $t$ is in years.

(a) Derive the relationship between the decay constant $k$ and the half-life $T_{1/2}$ from first principles. (3) (b) The isotope has a half-life of 8 years. Find $k$ to 3 s.f. (2) (c) What fraction of the original sample remains after 20 years? (3) (d) A different sample grows by continuous compounding with doubling time 5 hours. Derive its growth constant and state how long until it reaches 10× its initial size. (4)

Question 5 — Logarithmic scales, explain-out-loud (10 marks)

(a) The Richter magnitude is $M = \log_{10}(A/A_0)$ . Explain out loud what a jump from magnitude 5 to magnitude 7 means physically for the amplitude $A$ , and quantify it. (4) (b) pH is defined as $\text{pH} = -\log_{10}[\mathrm{H}^+]$ . If solution X has pH 3 and solution Y has pH 6, compute the ratio of their hydrogen-ion concentrations and explain why logarithmic scales are used here. (4) (c) Sound intensity level is $L = 10\log_{10}(I/I_0)$ dB. By how many dB does the level rise when intensity is multiplied by 100? (2)

Question 6 — Code-from-memory (6 marks)

Write a short pseudocode/Python function solve_growth(N0, k, target) that returns the time $t$ at which a continuously-growing quantity $N_0 e^{kt}$ first equals target. Show the underlying formula you inverted and briefly explain each line. (6)

Answer keyMark scheme & solutions

Question 1 (12)

(a) Product rule (4) Let $x=\log_a M$ and $y=\log_a N$ . By definition $a^x = M$ , $a^y = N$ . (1 — set up) Then $MN = a^x a^y = a^{x+y}$ by laws of exponents. (2 — key step) Taking $\log_a$ : $\log_a(MN) = x+y = \log_a M + \log_a N$ . (1 — conclude) ∎

(b) Power rule (4) Let $x=\log_a M$ , so $a^x = M$ . (1) Then $M^k = (a^x)^k = a^{kx}$ . (2) So $\log_a(M^k) = kx = k\log_a M$ . (1) ∎

(c) Change of base (4) Let $y = \log_a x$ , so $a^y = x$ . (1) Apply $\log_b$ to both sides: $\log_b(a^y) = \log_b x$ . (1) By the power rule: $y\log_b a = \log_b x$ . (1) Hence $y = \log_a x = \dfrac{\log_b x}{\log_b a}$ (valid since $\log_b a\neq 0$ ). (1) ∎

Question 2 (8)

(a) (4) $e = \lim_{n\to\infty}\left(1+\tfrac1n\right)^n$ . (1) Explanation: splitting a year into $n$ periods, each period's growth factor $1+\tfrac1n$ shrinks toward 1 while the number of compoundings $n$ grows. (1) The two effects offset: extra compoundings add smaller and smaller increments, so the product is bounded above and increasing, converging to a finite limit $\approx 2.71828$ (continuous compounding). (2)

(b) (4) — 1 mark each:

$n=1$ : $2^1 = 2$
$n=2$ : $1.5^2 = 2.25$
$n=4$ : $1.25^4 = 2.44141$
$n=12$ : $(1+1/12)^{12} = 2.61304$

Trend: values increase monotonically toward $e\approx 2.71828$ but stay below it. (comment mark included above)

Question 3 (12)

(a) (5) Take $\ln$ : $(2x-1)\ln5 = (x+2)\ln3$ . (1) $2x\ln5 - \ln5 = x\ln3 + 2\ln3$ . (1) $x(2\ln5 - \ln3) = 2\ln3 + \ln5$ . (1) $x = \dfrac{2\ln3 + \ln5}{2\ln5 - \ln3}$ . (1) Numerically $= \dfrac{2(1.0986)+1.6094}{2(1.6094)-1.0986} = \dfrac{3.8067}{2.1203} = 1.7954 \approx 1.80$ . (1)

(b) (4) Combine: $\log_2[x(x-2)] = 3 \Rightarrow x(x-2)=2^3=8$ . (1) $x^2 -2x -8 =0 \Rightarrow (x-4)(x+2)=0$ . (1) $x=4$ or $x=-2$ . (1) Reject $x=-2$ (needs $x>0$ and $x-2>0$ , i.e. domain $x>2$ ); so $x=4$ . (1)

(c) (3) Let $u=e^x$ : $u^2 -5u+6=0 \Rightarrow (u-2)(u-3)=0$ . (1) $e^x=2$ or $e^x=3$ . (1) $x=\ln2\approx0.693$ or $x=\ln3\approx1.099$ . (1)

Question 4 (12)

(a) (3) At $t=T_{1/2}$ , $N = N_0/2$ : $\tfrac12 N_0 = N_0 e^{-kT_{1/2}}$ . (1) $\tfrac12 = e^{-kT_{1/2}} \Rightarrow -kT_{1/2} = \ln\tfrac12 = -\ln2$ . (1) $\therefore k = \dfrac{\ln2}{T_{1/2}}$ . (1)

(b) (2) $k = \ln2/8 = 0.0866$ yr⁻¹. (2)

(c) (3) $N/N_0 = e^{-kt} = e^{-0.0866\times20}=e^{-1.733}$ . (1) $=0.1767$ . (1) So $\approx 17.7\%$ (equivalently $2^{-20/8}=2^{-2.5}=0.1768$ ). (1)

(d) (4) Doubling: $2 = e^{k\cdot5}\Rightarrow k=\ln2/5 = 0.1386$ hr⁻¹. (2) Reach 10×: $10 = e^{kt}\Rightarrow t = \ln10/k = \dfrac{\ln10}{\ln2/5} = 5\log_2 10 = 16.6$ hr. (2)

Question 5 (10)

(a) (4) $M=\log_{10}(A/A_0)$ , so $A/A_0 = 10^M$ . (1) Going 5→7 raises the exponent by 2, so $A$ increases by $10^{7-5}=10^2 = 100$ times. (2) Physically each unit of magnitude is a 10× amplitude jump; the scale compresses a huge amplitude range into small numbers. (1)

(b) (4) $[\mathrm H^+]=10^{-\text{pH}}$ . (1) $\dfrac{[\mathrm H^+]_X}{[\mathrm H^+]_Y}=\dfrac{10^{-3}}{10^{-6}}=10^{3}=1000$ . (2) Log scale used because concentrations span many orders of magnitude; a log axis makes them comparable. (1)

(c) (2) $\Delta L = 10\log_{10}(100)=10\times2 = 20$ dB. (2)

Question 6 (6)

Invert $N_0 e^{kt} = \text{target}$ : $t = \dfrac{1}{k}\ln\!\left(\dfrac{\text{target}}{N_0}\right)$ . (2 — formula)

import math
def solve_growth(N0, k, target):
    # N0 e^{kt} = target  ->  t = (1/k) ln(target/N0)
    if N0 <= 0 or target <= 0 or k == 0:
        raise ValueError("need positive N0, target and nonzero k")
    return math.log(target / N0) / k

(4 — correct inversion 2, guard/return 1, explanation of each line 1)

[
  {"claim":"Q3a solution x ≈ 1.795", "code":"x=(2*ln(3)+ln(5))/(2*ln(5)-ln(3)); result = abs(float(x)-1.7954) < 0.01"},
  {"claim":"Q3b: x=4 satisfies log2(x)+log2(x-2)=3", "code":"result = simplify(log(4,2)+log(2,2)-3)==0"},
  {"claim":"Q4c fraction remaining after 20yr ≈ 0.1768", "code":"k=ln(2)/8; frac=exp(-k*20); result = abs(float(frac)-0.1768) < 0.001"},
  {"claim":"Q4d time to reach 10x ≈ 16.6 hr", "code":"k=ln(2)/5; t=ln(10)/k; result = abs(float(t)-16.6096) < 0.01"},
  {"claim":"Q5c dB increase for 100x intensity is 20", "code":"dL=10*log(100,10); result = float(dL)==20"}
]