This gives the following graph.(b) The graph of the functionh(x) =ex/2isobtained by scaling the graph of thefunctionf(x) =exhorizontally by afactor of 2.This gives the following graph.15

Exercise Booklet 3Solution to Exercise 35(a) Taking the natural logarithm of bothsides gives3x= 70ln(3x) = ln 70xln 3 = ln 70x=ln 70ln 3= 3.87 (to 3 s.f.).(An alternative method is to take thelogarithm to base 3 of each side, whichgivesx= log370 = 3.87 (to 3 s.f.).)(b) Since 81 = 92= 34, you can immediatelysolve the equation to obtainx= 4.(The approach in part (a) would givex=ln 81ln 3= 4.)Solution to Exercise 36ln(12xy3)= ln12+ lnx+ ln(y3)(a)(by the first law)= ln(2−1)+ lnx+ ln(y3)=−ln 2 + lnx+ 3 lny(by the third law)= lnx+ 3 lny−ln 2ln3x2−12e2x(b)= ln(3x2−12)−ln(e2x)(by the second law)= ln(3(x2−4))−2x= ln(3(x−2)(x+ 2))−2x= ln 3 + ln(x−2) + ln(x+ 2)−2x(by the first law)Solution to Exercise 37(a) Letf(t) =aekt, whereaandkareconstants.We know thatf(6) = 500 andf(8) = 5000, soae6k= 500(1)ae8k= 5000.(2)Dividing equation (2) by equation (1)givesae8kae6k=5000500e8k−6k= 10e2k= 102k= ln 10k=12ln 10 = 1.1512925465. . .Equation (1) can be written asa(e2t)3= 500,and substitutinge2k= 10 (from themanipulation above) into this equationgivesa×103= 500a= 0.5.Therefore the functionfis given byf(t) = 0.5e0.5(ln 10)t(2≤t≤12);that is, approximately byf(t) = 0.5e1.151t(2≤t≤12).(b) The expected number of bacteria pergram after 12 hours isf(12) = 0.5e0.5 ln 10×12= 0.5e6 ln 10= 0.5(eln 10)6= 0.5×106= 500 000.(c) Every hour the number of bacteria ismultiplied bye(1/2) ln 10=(eln 10)1/2= 101/2=√10= 3.162 (to 4 s.f.).16

Solutions to exercisesSolution to Exercise 38(a) This inequality is true, as it is obtainedby multiplying both sides of the originalinequalityx >2 by the positivenumber 4.(b) This inequality is true, as it is obtainedby subtracting 2 from each side of theoriginal inequalityx >2.(c) This inequality is true, as it is obtainedby multiplying both sides of the originalinequality by the positive number12,(d) This inequality is false. Multiplying theoriginal inequality by−1 gives−x <−2,and it follows that the inequality−x >−2 is false.(e) This inequality is false. Multiplying theoriginal inequality by−12gives−x2<−1,and it follows that the inequality−x2>−1 is false.(f)It is not possible to say whether theinequalityax >2ais true or false, as wedo not know whetherais positive ornegative.Solution to Exercise 395x+ 2<2x+ 11(a)3x+ 2<113x <9x <3.The solution set is (−∞,3).5x−2≤2x−11(b)3x−2≤ −113x≤ −9x≤ −3.The solution set is (−∞,−3].x−23>1−x2(c)x−2>3−3x25x2>5x >2.The solution set is (2,∞).Solution to Exercise 40(a) Rearranging the inequality givesx2−x >6x2−x−6>0(x−3)(x+ 2)>0.The graph off(x) =x2−x+ 6 isu-shaped, withx-intercepts 3 and−2.So the solution set of the inequality is(−∞,−2)∪(3,∞).(Alternatively, instead of considering thegraph, you can find the solution set byconstructing a table of signs, as below.)x(−∞,−2)−2(−2,3)3(3,∞)x−3−−−0+x+ 2−0+++(x−3)×(x+ 2)+0−0+(b) Rearranging the inequality gives−2x2<2x2>−1.Asx2is always positive, the solution set

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