# 3 2 the final answer ln 3 ln 2 is just as acceptable

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32)(The final answer ln 3ln 2 is just asacceptable.)ln(4x) + 3 lnxln(e6) = ln(4x) + lnx36(v)= ln(4x×x3)6= ln(4x4)6(Other acceptable final answers includeln 4 + ln(x4)6 and ln 4 + 4 lnx6.)(vi)12ln(u8) = ln(u4)(The answer 4 lnuis just as acceptable.)(c) The pattern can be explained as follows. Itfollows from the logarithm laws that, for anyvalue ofn,log10(37×10n) = log1037 + log10(10n)= log1037 +n.So if you multiply 37 by 10n, then its commonlogarithm is increased byn.1567×2786 =eln(1567×2786)(d)=eln 1567+ln 2786e7.356 918 242+7.932 362 154e15.289 280 44 365 662(This is the exact answer.)Solution to Activity 475x= 0.5(a)ln(5x) = ln 0.5xln 5 = ln 0.5x=ln 0.5ln 5x=0.430 676. . .The solution isx=0.431 (to 3 s.f.).(Alternatively, you can proceed as follows:5x= 0.5x= log5(0.5)x=0.430 676. . . .)4e7t= 64(b)e7t= 16ln(e7t) = ln 167t= ln 16t=17ln 16t= 0.396 084. . .The solution ist= 0.396 (to 3 s.f.).313
Unit 3Functions5×2u/2+ 30 = 600(c)5×2u/2= 5702u/2= 114ln(2u/2) = ln 11412uln 2 = ln 114uln 2 = 2 ln 114u=2 ln 114ln 2u= 13.665 780. . .The solution isu= 13.7 (to 3 s.f.).(Alternatively, you can proceed as follows fromthe third equation above:2u/2= 114u2= log2114u= 2 log2114u= 13.665 780. . . .)23x5= 100(d)ln(23x5) = ln 100(3x5) ln 2 = ln 1003x5 =ln 100ln 23x=ln 100ln 2+ 5x=13ln 100ln 2+ 5x= 3.881 285 396. . .The solution isx= 3.88 (to 3 s.f.).(Alternatively, you can proceed as follows:23x5= 1003x5 = log2(100)3x= log2(100) + 5x=13(log2(100) + 5)x= 3.881 285 396. . . .)Solution to Activity 49Since ln 3 = 1.098 612 (to 7 s.f.), the rule offcan bewritten, approximately, asf(x) =e1.098 612x.Using the original form of the rule givesf(1.5) = 31.5= 5.20 (to 3 s.f.).Using the alternative form givesf(1.5) =e1.098 612×1.5= 5.20 (to 3 s.f.).Solution to Activity 50(a) Letf(t) =aekt, whereaandkare constants.Thenf(9) = 300 andf(12) = 4200, soae9k= 300andae12k= 4200.(5)Henceae12kae9k=4200300,which givese12k9k= 14e3k= 143k= ln 14k=13ln 14k= 0.879 685. . . .The first of equations (5) can be written asa(e3k)3= 300 and substitutinge3k= 14 intothis equation givesa×143= 300,soa=300143=75686= 0.109 329. . . .Soa= 0.109 andk= 0.880, both to threesignificant figures.Hence the required functionfis given,approximately, byf(t) = 0.109e0.880t(8t24).(b) The predicted number of bacteria per millilitreafter 24 hours isf(24) = (0.109 329. . .)e(0.879 685...)×24= 1.6×108(to 2 s.f.).Solution to Activity 51(a) Every decade the size of the tree population ispredicted to multiply by the factore0.06×1= 1.06 (to 3 s.f.).(b) Every century (10 decades) the size of the treepopulation is predicted to multiply by the factore0.06×10= 1.82 (to 3 s.f.).(c) Every five years (0.5 decades) the size of thetree population is predicted to multiply by thefactore0.06×0.5= 1.03 (to 3 s.f.).314
Solutions to activitiesSolution to Activity 52(a) Every year the level of radioactivity is predictedto multiply by the factore0.035×1= 0.97 (to 2 s.f.).(b) Every 25 years the level of radioactivity ispredicted to multiply by the factore0.035×25= 0.42 (to 2 s.f.).

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