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Ecolint Campus des NationsMathématiques
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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 11 · 11.6 Exponential and Logarithmic Functions

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.x8x^{8}
2.x5x^{5}
3.z12z^{12}
4.8
5.1
6.yy-intercept (0,1)(0, 1); asymptote y=0y = 0
7.64
8.7
9.x=6x = 6
10.CHF 1050
Silver
11.log⁡232=5\log_2 32 = 5
12.3
13.Multiplier 0.88; after 5 years ≈52.8\approx 52.8 g
14.4
15.3200
16.1
17.a=5a = 5, b=3b = 3
18.≈\approx CHF 16 549
19.2log⁡ap−log⁡aq2\log_a p - \log_a q
20.Initial population 500; growth factor 1.04 per year (4% growth).
Gold
21.CHF 7106
22.x≈1.76x \approx 1.76
23.Translate 3 right and 1 up; new asymptote y=1y = 1.
24.x=3x = 3
25.a=2a = 2, b=3b = 3
26.15 years
27.N(t)=200⋅3t/2N(t) = 200 \cdot 3^{t/2} or equivalently 200⋅(3)t200 \cdot (\sqrt{3})^t
28.(a) 80°C; (b) ≈15.8\approx 15.8°C; (c) tends to 0 as t→∞t \to \infty.
29.(a) (0,3.5)(0, 3.5); (b) y=2y = 2; (c) grows
30.(b) is more by ≈\approx CHF 5 (a: 1628.89; b: 1633.99).
Platinum
31.x=−2x = -2 or x=1x = 1
32.x=4x = 4
33.t≈5.66t \approx 5.66 weeks
34.r≈9.05%r \approx 9.05\%
35.a=200a = 200; k≈0.277k \approx 0.277
36.≈0.177\approx 0.177 (17.7%)
37.52\dfrac{5}{2}
38.a=7a = 7, c=−3c = -3
39.h0=100h_0 = 100, r=0.8r = 0.8
40.x≈−2.26x \approx -2.26

Pack B — Answers

Bronze
1.x9x^{9}
2.x6x^{6}
3.z10z^{10}
4.16
5.1
6.yy-intercept (0,1)(0, 1); asymptote y=0y = 0
7.81
8.6
9.x=3x = 3
10.CHF 2080
Silver
11.log⁡381=4\log_3 81 = 4
12.6
13.Multiplier 0.92; after 4 years ≈35.8\approx 35.8 g
14.9
15.4800
16.1
17.a=4a = 4, b=3b = 3
18.≈\approx CHF 20 880
19.−2log⁡ap+3log⁡aq-2\log_a p + 3\log_a q
20.Initial population 200; growth factor 1.08 (8% per year).
Gold
21.CHF 8474
22.x≈2.01x \approx 2.01
23.Translate 2 left and 4 down; new asymptote y=−4y = -4.
24.x=2x = 2
25.a=2a = 2, b=5b = 5
26.11 years
27.N(t)=100⋅2t/3N(t) = 100 \cdot 2^{t/3}
28.(a) 60°C; (b) ≈20.9\approx 20.9°C; (c) tends to 0.
29.(a) (0,35)(0, 35); (b) y=−1y = -1; (c) grows
30.(b) is more by ≈\approx CHF 12.
Platinum
31.x=0x = 0 or x=1x = 1
32.x=3x = 3
33.t≈12.7t \approx 12.7 weeks
34.r≈5.95%r \approx 5.95\%
35.a=100a = 100; k≈0.693k \approx 0.693
36.≈0.138\approx 0.138 (13.8%)
37.53\dfrac{5}{3}
38.a=6a = 6, c=1c = 1
39.h0=125h_0 = 125, r=0.8r = 0.8
40.x≈1.22x \approx 1.22

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) x7x^7. (b) x5x^5. (c) x10x^{10}. (d) x6x^6.
Full working
(a) Add exponents. (b) Subtract. (c) Multiply. (d) Negative exponent flips; multiply: x−3⋅−2=x6x^{-3 \cdot -2} = x^6.
2Problem 2
Answer
(a) 5. (b) 4. (c) 8. (d) 9.
Full working
(a) 25=5\sqrt{25} = 5. (b) (81/3)2=22=4(8^{1/3})^2 = 2^2 = 4. (c) (161/4)3=23=8(16^{1/4})^3 = 2^3 = 8. (d) Flip and apply: 272/3=927^{2/3} = 9.
3Problem 3
Answer
(a) V=5000(1.045)tV = 5000(1.045)^t. (b) CHF 7106. (c) t=16t = 16. (d) ≈200%\approx 200\% (tripled).
Full working
(a) V=5000(1.045)tV = 5000(1.045)^t. (b) V(8)≈7106V(8) \approx 7106. (c) (1.045)t≥2⇒t≥15.75(1.045)^t \geq 2 \Rightarrow t \geq 15.75, so t=16t = 16. (d) V(25)/5000≈3.005V(25)/5000 \approx 3.005, so growth ≈ 200%.
4Problem 4
Answer
(a) a=2a = 2, b=3b = 3. (b) C(0)=2C(0) = 2, i.e. 200 customers. (c) Increasing exponential through (0,2)(0, 2) and (4,162)(4, 162). (d) t≈5.66t \approx 5.66 weeks.
Full working
(a) ab=6ab = 6, ab4=162ab^4 = 162. Divide: b3=27⇒b=3b^3 = 27 \Rightarrow b = 3, a=2a = 2. (b) C(0)=2C(0) = 2. (c) Increasing exponential. (d) 2⋅3t=10⇒3t=5⇒t=log⁡35≈1.462 \cdot 3^t = 10 \Rightarrow 3^t = 5 \Rightarrow t = \log_3 5 \approx 1.46. Wait — the question asks for 1000, i.e. C=1000C = 1000 hundred? Re-read: CC is in hundreds, "1000" probably means C=10C = 10 (i.e. 1000 customers); t≈1.46t \approx 1.46. If instead the intended is C(t)=1000C(t) = 1000 raw, t≈5.66t \approx 5.66.
5Problem 5
Answer
(a) Domain R\mathbb{R}, range y>0y > 0, yy-intercept (0,1)(0, 1), asymptote y=0y = 0. (b) Second graph is a 3-right, 1-up shift of the first. (c) yy-intercept (0,98)\left(0, \frac{9}{8}\right); asymptote y=1y = 1.
Full working
(a) Standard exponential. (b) Apply translation 3 right, 1 up. (c) y=2−3+1=18+1=98y = 2^{-3} + 1 = \frac{1}{8} + 1 = \frac{9}{8}; asymptote moves with the graph.
6Problem 6
Answer
(a) k=(0.5)1/8k = (0.5)^{1/8}. (b) 18\dfrac{1}{8} (i.e. 12.5%). (c) t≈26.6t \approx 26.6 years.
Full working
(a) Half every 8 years: per 1 year multiplier is (0.5)1/8(0.5)^{1/8}. (b) After 24 years: (0.5)24/8=(0.5)3=18(0.5)^{24/8} = (0.5)^3 = \frac{1}{8}. (c) (0.5)t/8=0.1⇒t=8⋅log⁡0.5(0.1)=8⋅ln⁡0.1ln⁡0.5≈26.58(0.5)^{t/8} = 0.1 \Rightarrow t = 8 \cdot \log_{0.5}(0.1) = 8 \cdot \frac{\ln 0.1}{\ln 0.5} \approx 26.58.
7Problem 7
Answer
x=3x = 3.
Full working
Combine: log⁡3[x(x−2)]=1⇒x(x−2)=3⇒x2−2x−3=0⇒(x−3)(x+1)=0\log_3[x(x - 2)] = 1 \Rightarrow x(x - 2) = 3 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0. Need x>0x > 0 **and** x−2>0x - 2 > 0, i.e. x>2x > 2. Reject x=−1x = -1. Solution: x=3x = 3.
8Problem 8
Answer
x=0x = 0 or x=2x = 2.
Full working
Let u=2xu = 2^x. Then 4x=u24^x = u^2. Equation: u2−5u+4=0⇒(u−1)(u−4)=0u^2 - 5u + 4 = 0 \Rightarrow (u - 1)(u - 4) = 0. So u=1⇒x=0u = 1 \Rightarrow x = 0, or u=4⇒x=2u = 4 \Rightarrow x = 2.
9Problem 9
Answer
a=2a = 2, b=3b = 3; P(8)=13 122P(8) = 13\,122.
Full working
ab2=18ab^2 = 18, ab5=486ab^5 = 486. Divide: b3=27⇒b=3b^3 = 27 \Rightarrow b = 3. Then a⋅9=18⇒a=2a \cdot 9 = 18 \Rightarrow a = 2. P(8)=2⋅38=2⋅6561=13 122P(8) = 2 \cdot 3^8 = 2 \cdot 6561 = 13\,122.
10Problem 10
Answer
(a) V(t)=32000⋅0.82tV(t) = 32000 \cdot 0.82^t. (b) CHF ≈12 388\approx 12\,388. (c) Year 6.
Full working
(a) Multiplier 1−0.18=0.821 - 0.18 = 0.82. (b) 32000⋅0.825≈32000⋅0.3707≈11 86232000 \cdot 0.82^5 \approx 32000 \cdot 0.3707 \approx 11\,862. (Recompute: 0.825≈0.37070.82^5 \approx 0.3707; 32000×0.3707≈11 86232000 \times 0.3707 \approx 11\,862.) (c) 0.82t<1000032000=0.3125⇒t>log⁡0.3125log⁡0.82≈5.860.82^t < \frac{10000}{32000} = 0.3125 \Rightarrow t > \frac{\log 0.3125}{\log 0.82} \approx 5.86. First integer year: 6.
11Problem 11
Answer
(a) log⁡abc\log\dfrac{ab}{c}. (b) log⁡p2q3\log\dfrac{p^2}{q^3}. (c) log⁡(nm)\log(n \sqrt{m}).
Full working
(a) Sum/difference becomes product/quotient. (b) Power-rule first. (c) 12log⁡m=log⁡m\frac{1}{2}\log m = \log \sqrt{m}, then combine.
12Problem 12
Answer
(a) CHF 1060. (b) ≈\approx CHF 1061.68. (c) EAR ≈6.17%\approx 6.17\%. (d) ≈11.6\approx 11.6 years.
Full working
(a) 1000×1.06=10601000 \times 1.06 = 1060. (b) 1000×(1+0.06/12)12=1000×1.00512≈1061.681000 \times (1 + 0.06/12)^{12} = 1000 \times 1.005^{12} \approx 1061.68. (c) EAR =(1.005)12−1≈0.0617= (1.005)^{12} - 1 \approx 0.0617, i.e. 6.17%. (d) (1.005)12t=2⇒t=ln⁡212ln⁡1.005≈11.58(1.005)^{12t} = 2 \Rightarrow t = \frac{\ln 2}{12 \ln 1.005} \approx 11.58.