Exponentials & Logarithms

Question 1

(a) Half-life is time for $N$ to halve: $N(T_{1/2}) = \tfrac12 N_0$. (1) So $N_0 e^{-\lambda T_{1/2}} = \tfrac12 N_0 \Rightarrow e^{-\lambda T_{1/2}} = \tfrac12$. (1) Take $\ln$: $-\lambda T_{1/2} = \ln\tfrac12 = -\ln 2 \Rightarrow T_{1/2} = \dfrac{\ln 2}{\lambda}$. (1)

(b) $\lambda = \dfrac{\ln 2}{5730}$. From $0.63 = e^{-\lambda t}$: $t = -\dfrac{\ln 0.63}{\lambda} = \dfrac{5730 \ln(1/0.63)}{\ln 2}$. (2) $= \dfrac{5730 \times 0.46204}{0.69315} = 5730 \times 0.66658 = 3819.5$. (1) Age $\approx \mathbf{3819}$ years (nearest year). (1)

(c) After $n$ steps, $P = P_0(1+r/n)^n$. (1) Write $(1+r/n)^n = \left[(1+\tfrac{1}{n/r})^{n/r}\right]^{r}$. (1) Let $m=n/r\to\infty$ as $n\to\infty$; the inner bracket $\to e$ by definition $e=\lim_{m\to\infty}(1+\tfrac1m)^m$. (1) Hence $P \to P_0 e^{r}$. (1)

(d) $P_{\text{loop}}(4)/P_0 = (1+0.02)^4 = 1.02^4 = 1.08243216$. (1) $e^{0.08} = 1.0832871$. Relative error $= \dfrac{1.0832871 - 1.0824322}{1.0832871} = 7.892\times10^{-4}$. (1) $= \mathbf{0.0789\%}$ (3 s.f.). (1)

(e) Doubling time $T_d = \dfrac{\ln 2}{r} = \dfrac{0.69315}{0.08} = 8.664$ years $\approx \mathbf{8.66}$ years. (2) From $2P_0 = P_0 e^{rT_d}$, $P_0$ cancels, so $T_d$ depends only on $r$, not $P_0$ — the fractional growth rate is scale-free. (2)

Question 2

(a) Let $m=\log_a x$, $n=\log_a y$, so by definition $a^m = x$, $a^n = y$. (1) Then $xy = a^m a^n = a^{m+n}$ by the law of exponents. (1) Applying $\log_a$ (inverse of $a^{(\cdot)}$): $\log_a(xy) = m+n$. (1) $= \log_a x + \log_a y$. (1)

(b) Let $y=\log_a x$, so $a^y = x$. Take $\log_b$ of both sides: $\log_b(a^y)=\log_b x$. (1) Power law: $y\log_b a = \log_b x$. (1) So $y = \dfrac{\log_b x}{\log_b a}$, i.e. $\log_a x = \dfrac{\log_b x}{\log_b a}$. (1)

(c) Put $x=b$ in (b): $\log_a b = \dfrac{\log_b b}{\log_b a} = \dfrac{1}{\log_b a}$. (2) Hence $\log_a b \cdot \log_b a = 1$, provided $a,b>0$ and $a,b\neq 1$. (1)

(d) $2\log_3 x - \log_3(x-2) = 2 \Rightarrow \log_3\dfrac{x^2}{x-2} = 2$. (1) So $\dfrac{x^2}{x-2} = 3^2 = 9$. (1) $x^2 = 9x - 18 \Rightarrow x^2 - 9x + 18 = 0$. (1) $(x-3)(x-6)=0 \Rightarrow x=3$ or $x=6$. (1) Domain requires $x>0$ and $x-2>0$, i.e. $x>2$. (1) Both $3$ and $6$ satisfy this, so $x=3$ and $x=6$ are both valid. (1)

Question 3

(a) $M_1 - M_2 = \log_{10}\dfrac{A_1}{A_0} - \log_{10}\dfrac{A_2}{A_0} = \log_{10}\dfrac{A_1}{A_2}$. (1) $7.2-5.8 = 1.4 = \log_{10}(A_1/A_2)$, so $A_1/A_2 = 10^{1.4}$. (1) $10^{1.4} = 25.1189$. (1) Factor $\approx \mathbf{25.1}$. (1)

(b) $\Delta L = 10\log_{10}\dfrac{2I}{I} = 10\log_{10}2$ (exact). (2) $= 10 \times 0.30103 = 3.0103 \approx \mathbf{3.01\ dB}$. (2)

(c) Take $\ln$: $(2x+1)\ln 3 = (x-2)\ln 5$. (1) $2x\ln3 + \ln3 = x\ln5 - 2\ln5 \Rightarrow x(2\ln3 - \ln5) = -2\ln5 - \ln3$. (1) $x = \dfrac{-2\ln5 - \ln3}{2\ln3 - \ln5} = \dfrac{-(\ln 75)}{\ln(9/5)}$. (1) $= \dfrac{-4.31749}{0.58779} = \mathbf{-7.346}$ (4 s.f.). (1)

(d) $[\mathrm{H}^+]_A = 10^{-3.4}$. $[\mathrm{H}^+]_B = 5\times10^{-3.4}$. (1) $\text{pH}_B = -\log_{10}(5\times10^{-3.4}) = 3.4 - \log_{10}5$. (2) $= 3.4 - 0.69897 = 2.701 \approx \mathbf{2.70}$. (1)

[
  {"claim":"C-14 artefact age ~3819 years",
   "code":"t = 5730*ln(1/Rational(63,100))/ln(2); result = abs(float(t)-3819) < 1"},
  {"claim":"Quarterly relative error ~0.0789%",
   "code":"e_true = exp(Rational(8,100)); loop = (1+Rational(2,100))**4; rel = (e_true-loop)/e_true; result = abs(float(rel)*100 - 0.0789) < 0.001"},
  {"claim":"Q2d solutions are x=3 and x=6",
   "code":"sol = solve(Eq(x**2 - 9*x + 18, 0), x); result = set(sol) == {3,6}"},
  {"claim":"Q3c solution x ~ -7.346",
   "code":"xval = (-2*ln(5)-ln(3))/(2*ln(3)-ln(5)); result = abs(float(xval)-(-7.346)) < 0.001"},
  {"claim":"Q3d pH of B ~2.70",
   "code":"pHB = Rational(34,10) - log(5,10); result = abs(float(pHB)-2.70) < 0.01"}
]