"Essential Mathematics & Statistics for Science" by Dr G Currell & Dr A A Dowman (John Wiley & Sons) Answers to In-Text Questions

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1 "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 5 ii) (10 2 ) 3 = = 10 6 iii) = = 10 1 = 10 iv) = 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) = iii) 10-3 = 1/1000 = iv) e 4.2 = v) e 0 = 1 vi) e 1 = Q5.3 i) log(348) = ii) log(34.8) = iii) log(3.48) = iv) log(100) = 2 v) log(0.01) = -2 vi) log(0.5) = vii) log(2) = viii) log(20) = 1.301

2 ix) ln(e 1 ) = 1 x) ln(10) = Q5.4 i) e x =22 x = ln(22) = ii) e 3x =22 3x = ln(22) = x = /3 = iii) 2e 3x =22 e 3x = 22/2 = 11 3x = ln(11) = x = /3 = iv) 10 x = 18 x = log(18) = v) 10 2x = 18 2x = log(18) = x = /2 = vi) x = x = 18/1.2 = 15 2x = log(15) = x = /2 = Q5.5 i) log( ) = -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) = = 1.3 v) log(200) = log(2 100) = log(2) + log(100) = 2.3 vi) log(0.5) = log(1 2) = log(1) - log(2) = = vii) ln(1000) = 2.30 log(1000) = = 6.90 viii) ln(2) = 2.30 log(2) = = 0.69 ix) log(2 3.1 ) = 3.1 log(2) = = 0.93 x) ln( ) = ln(0.407) + ln(7.39) = = 1.1 Q5.6 i) e p := ln(e p ) = ln(0.006) = p = ii) 10 4p = 3.0

3 log(10 4p ) = 4p = log(3.0) 4p = p = iii) e -2p = ln(e -2p ) = -2p = ln(0.006) = 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 = / 4 = 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) = / = 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 = = 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 = ( ) - ( ) = ii) Difference in [H+] between strong acid and pure water = ( ) - ( ) = iii) No iv) Ratio of [H+] between pure water and strong base = ( )/( ) = v) Ratio of [H+] between strong acid and pure water = ( )/( ) = 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

4 Q5.9 i) ph = - log( ) = 8.47 ii) 4.2 = -log[h + ] Hence log[h + ] = [H + ] = = mol L -1. Q5.10 T% A Percentage Transmittance Absorbance 0% (infinity) 0.1% 3 1% 2 10% 1 50% % 0 Q5.11 The remaining money will fall as follows: End of Day n i) halving Money, M( ) ii) using formula Typical calculation for (ii) Money, M( ) If n=4 M = 640 exp( ) = 640 exp(-2.772) = = 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.

5 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 The first step is to solve the following equation for n = 6 n Taking logs of both sides (see 5.1 Mathematics of e, log & ln): log( ) = log(6 n ) 7.30 = n log(6) = n Hence: n = 7.30 / = 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 = = ii) iii) g = 0.5 is equivalent to a 50% drop for each period. The equation becomes: A n = (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 = (0.5) 5.5 = Q5.14 i) If t = 25 days, N 25 = 3500 exp( ) = 3500 exp(0.5)= = 5771 ii) If t = 50 days, N 50 = 3500 exp( ) = 3500 exp(1)= = 9514 iii) If t = 75 days, N 75 = 3500 exp( ) = 3500 exp(1.5)= = 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.

6 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) = Hence our equation becomes: N t = 100 e ( t) = 100exp( 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 = = 450 exp(k 10) Dividing both sides by 450: exp(k 10) = 620/450 = Taking natural logs of both sides; ln(exp(k 10)) = ln(1.378) = ln(exp(k 10)) = k 10 Hence k = / 10 = The complete equation is therefore: N t = 450 exp( t) ii) Substituting for t = 12 gives: N 12 = 450 exp( ) = 661 Q5.17 Using the log expression of [5.28] t 180 G = = log N / N log / = 28 minutes ( ) t 0 4 ( ) = 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( /20) = = We need not have calculated this as by definition the population doubles in the generation time ii) N 60 = 1000 exp( /20) = = Again, the population doubles every 20 minutes, so after 60 minutes it has

7 doubled 3 times, = 8 times increase. iii) N 10 = 1000 exp( /20) = = iv) N 45 = 1000 exp( /20) = = Q5.19 i) 3 days = 1 half-life. Fraction left = 1/2 = 0.5 ii) 6 days = 2 half-life. Fraction left = = 0.25 = 1/4 iii) 9 days = 3 half-life. Fraction left = = = 1/8 Q5.20 We will use the equation: A t = A O exp( 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( /T) = exp( / T) To solve this equation, we take natural logs of both sides: ln(0.1) = ln(exp( / T)) giving = / T Hence, the half-life, T, is given by T = / = 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 = = Hence we can write 0.01 = exp{-t/( )} To solve this equation, we take natural logs of both sides: ln(0.01) = ln(exp{-t/( )}) giving = -t/( ) Hence, the time, t, is given by t = = s = 15.2 ms Q5.22 i) See also In-Text questions, Other Files QE5

8 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 V V t 4 ii) Reading from graph for t = T = 1.5 gives V 6.5 so V/V t = T = 3.0 gives V 8.5 so V/V iii) Substituting into the equation for : t = T = 1.5 gives V = 6.32 so V/V and t = 2T = 3.0 gives V = 8.65 so V/V