Exponentials & Logarithms

Question 1 (10)

(a) Losing $18\%$ leaves $82\%$: $N_0 e^{-5k} = 0.82 N_0$. (1) $e^{-5k} = 0.82 \Rightarrow -5k = \ln 0.82$ (1) $k = -\dfrac{\ln 0.82}{5} = \dfrac{0.198451}{5} = 0.039690$ (1) $k \approx 0.03969$ (4 s.f.) (1) Why: the fraction remaining, not lost, drives the decay equation.

(b) Half-life: $e^{-k t_{1/2}} = \tfrac12$ (1) $t_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693147}{0.039690} = 17.46$ years (2)

(c) $T = 17.46$ years. (1) Both models describe the same physical decay; $2^{-t/T}$ halves every $T$ years, which is by definition the half-life. So $T = t_{1/2}$. (2) Why: $e^{-kt} = 2^{-t/T} \Rightarrow kt = (t/T)\ln 2 \Rightarrow T = \ln2/k$, identical to (b).

Question 2 (10)

(a) Take $\ln$: $(2x+1)\ln 3 = (x-2)\ln 5$ (1) $2x\ln3 + \ln3 = x\ln5 - 2\ln5$ $x(2\ln3 - \ln5) = -2\ln5 - \ln3$ (1) $x = \dfrac{-2\ln5 - \ln3}{2\ln3 - \ln5} = \dfrac{-(\ln 75)}{\ln(9/5)}$ (1) Numerically $x = \dfrac{-4.31749}{0.58779} = -7.346$ (1)

(b) $\log_2[(x-1)(x+2)] = 2 \Rightarrow (x-1)(x+2) = 4$ (1) $x^2 + x - 2 = 4 \Rightarrow x^2 + x - 6 = 0 \Rightarrow (x+3)(x-2)=0$ (1) $x = 2$ or $x = -3$; reject $x=-3$ (makes $\log_2(x-1)$ undefined). $x = 2$. (1)

(c) Let $u = e^x$: $u^2 - 5u + 6 = 0 \Rightarrow (u-2)(u-3)=0$ (1) $u = 2$ or $u = 3$ (1) $x = \ln 2$ or $x = \ln 3$ (both valid). (1)

Question 3 (10)

(a) Domain of $f$: $2x - 3 > 0 \Rightarrow x > \tfrac32$. (1) Range of $g$: $e^x > 0 \Rightarrow g(x) > 1$. (1)

(b) $y = \ln(2x-3) \Rightarrow e^y = 2x - 3 \Rightarrow x = \dfrac{e^y + 3}{2}$ (2) $f^{-1}(x) = \dfrac{e^x + 3}{2}$ (1) Domain of $f^{-1}$ = range of $f$ = all real $x$, i.e. $x \in \mathbb{R}$. (1)

(c) Sketch marks:

• $y=f(x)$: vertical asymptote $x = \tfrac32$, passing through $(2,0)$, increasing. (1)
• $y=f^{-1}(x)$: horizontal asymptote $y = \tfrac32$, passing through $(0,2)$. (1)
• Line of symmetry $y = x$. (1)
• Correct reflection/relative positions shown. (1)

Question 4 (10)

(a) $\log_{10}(A_P/A_0) = 6.4$, $\log_{10}(A_Q/A_0) = 4.9$. (1) $\dfrac{A_P}{A_Q} = 10^{6.4 - 4.9} = 10^{1.5}$ (2) $= 31.6228\ldots \approx 31.6$ times. (1)

(b) $\log_{10}E = 1.5M + 4.8$. Increase $M$ by 1: $\Delta(\log_{10}E) = 1.5$. (1) Ratio $= \dfrac{E_{M+1}}{E_M} = 10^{1.5}$ (1) $= 31.62 \approx 31.6$. ∎ (1)

(c) $\dfrac{E_P}{E_Q} = 10^{1.5(6.4-4.9)} = 10^{1.5 \times 1.5} = 10^{2.25}$ (2) $= 177.8 \approx 178$ times. (1)

Question 5 (10)

(a) Doubling: $P(t+T_d) = 2P(t)$ (1) $P_0 a^{t+T_d} = 2 P_0 a^{t} \Rightarrow a^{t}\cdot a^{T_d} = 2 a^{t}$ (1) Divide by $a^t$: $a^{T_d} = 2$. (1) Take $\ln$: $T_d \ln a = \ln 2 \Rightarrow T_d = \dfrac{\ln 2}{\ln a}$. ∎ (1)

(b) $P(3) = 500 a^3 = 2000 \Rightarrow a^3 = 4 \Rightarrow a = 4^{1/3} = 1.5874$. (1) $\ln a = \tfrac13\ln 4 = 0.462098$. (1) $T_d = \dfrac{\ln 2}{\ln a} = \dfrac{0.693147}{0.462098} = 1.5$ hours. (1)

(c) Change of base: $\dfrac{\ln 2}{\ln a} = \dfrac{\log_{10}2 / \log_{10}e}{\log_{10}a / \log_{10}e} = \dfrac{\log_{10}2}{\log_{10}a}$. (1) $= \dfrac{0.30103}{0.20069} = 1.5$ hours. (1) Same value — confirmed. (1)

[
  {"claim":"Q1(a) k = 0.03969 (4sf)","code":"k = -ln(Rational(82,100))/5; result = round(float(k),5)==0.03969"},
  {"claim":"Q1(b) half-life ~ 17.46 yr","code":"k = -ln(Rational(82,100))/5; th = ln(2)/k; result = round(float(th),2)==17.46"},
  {"claim":"Q2(a) x = -ln(75)/ln(9/5) ~ -7.346","code":"x = (-2*ln(5)-ln(3))/(2*ln(3)-ln(5)); result = round(float(x),3)==-7.346"},
  {"claim":"Q4(c) energy ratio = 10**2.25 ~ 178","code":"r = Integer(10)**Rational(9,4); result = round(float(r))==178"},
  {"claim":"Q5(b) a=4**(1/3), doubling time = 1.5 h","code":"a = Integer(4)**Rational(1,3); Td = ln(2)/ln(a); result = round(float(Td),4)==1.5"}
]