"Essential Mathematics & Statistics for Science" by Dr G Currell & Dr A A Dowman (John Wiley & Sons) Answers to In-Text Questions 5 Logarithmic & Exponential Functions To navigate, use the Bookmarks in the PDF file Q5.1 i) 10 2 10 3 = 10 2+3 = 10 5 ii) (10 2 ) 3 = 10 2 3 = 10 6 iii) 10 3 10 2 = 10 3-2 = 10 1 = 10 iv) 10 3 10-2 = 10 3-(-2) = 10 3+2 = 10 5 v) e 2 e 0 = e 2+0 = e 2 vi) e 3 e 2 = e 3+2 = e 5 vii) e 5 e 3 = e 5-3 = e 2 viii) (e 2 ) 3 = e 2 3 = e 6 Q5.2 i) 10 3 = 1000 ii) 10 3.4 = 2511.88... iii) 10-3 = 1/1000 = 0.001 iv) e 4.2 = 66.686... v) e 0 = 1 vi) e 1 = 2.718... Q5.3 i) log(348) = 2.541 ii) log(34.8) = 1.541 iii) log(3.48) = 0.541 iv) log(100) = 2 v) log(0.01) = -2 vi) log(0.5) = -0.301 vii) log(2) = 0.301 viii) log(20) = 1.301
ix) ln(e 1 ) = 1 x) ln(10) = 2.302 Q5.4 i) e x =22 x = ln(22) = 3.091 ii) e 3x =22 3x = ln(22) = 3.091 x = 3.091 /3 = 1.030 iii) 2e 3x =22 e 3x = 22/2 = 11 3x = ln(11) = 2.397 x = 2.397 /3 = 0.799 iv) 10 x = 18 x = log(18) = 1.255 v) 10 2x = 18 2x = log(18) = 1.255 x = 1.255 /2 = 0.627 vi) 1.2 10 2x = 18 10 2x = 18/1.2 = 15 2x = log(15) = 1.176 x = 1.176 /2 = 0.588.. Q5.5 i) log(10-0.3 ) = -0.3 ii) ln(e 0.62 ) = 0.62 iii) log(2) = 0.30 approx iv) log(20) = log(2 10) = log(2) + log(10) = 0.30 + 1 = 1.3 v) log(200) = log(2 100) = log(2) + log(100) 0.30 + 2 = 2.3 vi) log(0.5) = log(1 2) = log(1) - log(2) = 0-0.30 = -0.30 vii) ln(1000) = 2.30 log(1000) = 2.30 3 = 6.90 viii) ln(2) = 2.30 log(2) = 2.30 0.30 = 0.69 ix) log(2 3.1 ) = 3.1 log(2) = 3.1 0.30 = 0.93 x) ln(0.407 7.39) = ln(0.407) + ln(7.39) = -0.90 + 2 = 1.1 Q5.6 i) e p := 0.006 ln(e p ) = ln(0.006) = -5.12 p = -5.12 ii) 10 4p = 3.0
log(10 4p ) = 4p = log(3.0) 4p = 0.477 p = 0.119 iii) e -2p = 0.006 ln(e -2p ) = -2p = ln(0.006) = -5.12 p = 5.12/2 = 2.56 iv) 10-4p = 33.0 log(10-4p ) = -4 plog(10) = -4p 1 = log(33) p = -log(33)/4 = -1.52 / 4 = -0.38 v) 4 p = 33 log(33) = log(4 p ) = p log(4) 0.602p = 1.52 p = 2.52 vi) 0.2 p = 0.3 take log or ln in this case p log (0.2) = log(0.3) p = log(0.3)/ log (0.2) = -0.524/-0.699 = 0.748 Q5.7 Initial loudness = 70dB i) Doubling power density adds 3dB to loudness. New value = 73dB ii) Increasing by 8 (= 2 3 ) adds 3 3dB = 9dB. New value = 79dB iii) Increasing by 100: difference = 10 log(p 1 /P 2 ) = 10 log(100) = 10 2 = 20 New value = 70 + 20 = 90dB iv) Halving reduces the loudness by 3dB ( see (i)) New value = 67dB v) See (ii) subtract 9dB. New value = 61dB vi) See (iii). Subtract 20dB. New value = 50dB. Q5.8 i) Difference in [H+] between pure water and strong base = (1 10-7 ) - (1 10-13 ) = 9.99999 10-8 ii) Difference in [H+] between strong acid and pure water = (1 10-1 ) - (1 10-7 ) = 0.0999999 iii) No iv) Ratio of [H+] between pure water and strong base = (1 10-7 )/(1 10-13 ) = 1 10 6 v) Ratio of [H+] between strong acid and pure water = (1 10-1 )/(1 10-7 ) = 1 10 6 vi) Yes for this example of strong acid and base vii) Difference in ph between pure water and strong base = 7 13 = -6 viii) Difference in ph between strong acid and pure water = 1 7 = -6 ix) Yes for this example of strong acid and base
Q5.9 i) ph = - log(3.4 10-9 ) = 8.47 ii) 4.2 = -log[h + ] Hence log[h + ] = - 4.2 [H + ] = 10-4.2 = 6.31 10-5 mol L -1. Q5.10 T% A Percentage Transmittance Absorbance 0% (infinity) 0.1% 3 1% 2 10% 1 50% 0.30 100% 0 Q5.11 The remaining money will fall as follows: End of Day n 0 1 2 3 4 5 6 7 i) halving Money, M( ) 640 320 160 80 40 20 10 5 ii) using formula Typical calculation for (ii) Money, M( ) 640 320 160 80 40 20 10 5 If n=4 M = 640 exp(-0.693 4) = 640 exp(-2.772) = 640 0.0625 = 40.02 Which is close to 40. Q5.12 Using the equation: A n = A O g n Each week is one stage in the process, and we use n to denote the number of weeks. At each stage, the number of people actively involved will increase by a factor of '6'. Hence, the gain factor for this problem, g = 6. At the beginning, when n = 0, there is just one person involved, and we write, A O = 1.
Hence our equation becomes: A n = 1 6 n To solve the problem we need to find the value of n such that A n 20 10 6. The first step is to solve the following equation for n. 20 10 6 = 6 n Taking logs of both sides (see 5.1 Mathematics of e, log & ln): log(20 10 6 ) = log(6 n ) 7.30 = n log(6) = n 0.778 Hence: n = 7.30 / 0.778 = 9.4 weeks Thus we can see that by the end of 10 weeks, at least one third of the UK population would be involved - assuming of course that the chain is not broken (which thankfully it always is!). Q5.13 i) A period of 2 hours is equivalent to 5 'half-lives', where one half-life, T, = 0.4. The initial population will drop by 50% for each half-life. Population after 5 half-lives = 2.0 10 6 0.5 0.5 0.5 0.5 0.5 = 6.25 10 4 ii) iii) g = 0.5 is equivalent to a 50% drop for each period. The equation becomes: A n = 2.0 10 6 (0.5) n For a time, t = 2.2 hours, n = t/t = 2.2 / 0.4 = 5.5. Evaluating the equation, gives A n = 2.0 10 6 (0.5) 5.5 = 4.419 10 4 Q5.14 i) If t = 25 days, N 25 = 3500 exp(0.02 25) = 3500 exp(0.5)= 3500 1.649 = 5771 ii) If t = 50 days, N 50 = 3500 exp(0.02 50) = 3500 exp(1)= 3500 2.718 = 9514 iii) If t = 75 days, N 75 = 3500 exp(0.02 75) = 3500 exp(1.5)= 3500 4.482 = 15686 Q5.15 Using the equation: N t = N O g t Each week is one stage in the process, and we use t to denote the number of weeks. At each stage, the number of people actively involved will increase by a factor of '1.1'. Hence, the gain factor for this problem, g = 1.1. At the beginning, when t = 0, g 0 = 1 and N 0 = 100.
Hence we can write: 100 = N O We now need to convert this problem to an equation of the form: N t = N O e kt and we must have that g = e k or k = ln(g) which gives k = ln(1.1) = 0.0953 Hence our equation becomes: N t = 100 e (0.0953 t) = 100exp(0.0953 t) Q5.16 i) We can put time t = 0 when N 0 = 450, then: 450 = N O exp(k 0) = N O giving the equation: N t = 450 e kt = 450 exp(kt) We can then substitute the values, N t = 620 at time, t = 10. 620 = 450 exp(k 10) Dividing both sides by 450: exp(k 10) = 620/450 = 1.378 Taking natural logs of both sides; ln(exp(k 10)) = ln(1.378) = 0.320 ln(exp(k 10)) = k 10 Hence k = 0.320 / 10 = 0.0320 The complete equation is therefore: N t = 450 exp(0.0320 t) ii) Substituting for t = 12 gives: N 12 = 450 exp(0.0320 12) = 661 Q5.17 Using the log expression of [5.28] t 180 G = 0.301 = 0.301 6 log N / N log 1.7 10 / 2.0 10 = 28 minutes ( ) t 0 4 ( ) = 0.301 180 1.929 Q5.18 N 0 = 1000, G = 20 minutes so we can write using [5.27] N t = 1000 exp(0.693 t/20) i) N 20 = 1000 exp(0.693 20/20) = 1000 2 = 2000. We need not have calculated this as by definition the population doubles in the generation time ii) N 60 = 1000 exp(0.693 60/20) = 1000 8 = 8000. Again, the population doubles every 20 minutes, so after 60 minutes it has
doubled 3 times, 2 2 2 = 8 times increase. iii) N 10 = 1000 exp(0.693 10/20) = 1000 1.414 = 1414. iv) N 45 = 1000 exp(0.693 45/20) = 1000 4.755 = 4755. Q5.19 i) 3 days = 1 half-life. Fraction left = 1/2 = 0.5 ii) 6 days = 2 half-life. Fraction left = 0.5 0.5 = 0.25 = 1/4 iii) 9 days = 3 half-life. Fraction left = 0.5 0.5 0.5 = 0.125 = 1/8 Q5.20 We will use the equation: A t = A O exp(-0.693 t/t) We need to calculate the proportion of remaining activity, (A t / A O ), where (A t / A O ) =exp(-0.693t/t) We know that, after t = 26 hours: (A 26 / A O ) = 1/10 = 0.1 Hence we can write 0.1 = exp(-0.693 26/T) = exp(-18.018 / T) To solve this equation, we take natural logs of both sides: ln(0.1) = ln(exp(-18.018 / T)) giving -2.303 = -18.018 / T Hence, the half-life, T, is given by T = 18.018 / 2.303 = 7.825 hours Q5.21 We will use the equation: A t = A O exp(-t/τ) where A will represent the charge on the capacitor. The proportion of remaining charge, (A t / A O ), after time, t, will be given by (A t / A O ) = exp(-t/τ) We know that (A t / A O ) = 1% = 0.01 τ = CR = 0.01 10-6 330 10 3 = 3.3 10-3 Hence we can write 0.01 = exp{-t/(3.3 10-3 )} To solve this equation, we take natural logs of both sides: ln(0.01) = ln(exp{-t/(3.3 10-3 )}) giving -4.605 = -t/(3.3 10-3 ) Hence, the time, t, is given by t = 4.605 3.3 10-3 = 15.2 10-3 s = 15.2 ms Q5.22 i) See also In-Text questions, Other Files QE5
Using the equation: V = V 0 {1-exp(-t/T)} with T = 1.5, and V 0 = 10 values for V are calculated for values of t as below, and a graph drawn. t 0 1 2 3 4 V 0.00 4.87 7.36 8.65 9.31 10 V 8 6 4 2 0 0 1 2 3 t 4 ii) Reading from graph for t = T = 1.5 gives V 6.5 so V/V 0 0.65 t = T = 3.0 gives V 8.5 so V/V 0 0.85 iii) Substituting into the equation for : t = T = 1.5 gives V = 6.32 so V/V 0 0.632 and t = 2T = 3.0 gives V = 8.65 so V/V 0 0.865