Solving Logarithms and Exponential Equations

Transcription

1 Solvig Logarithms ad Epoetial Equatios Logarithmic Equatios There are two major ideas required whe solvig Logarithmic Equatios. The first is the Defiitio of a Logarithm. You may recall from a earlier topic: If a umber (N) is writte as a power with a base (b) ad a epoet (e), such as e N = b the log b N = e. The base, b, must be a positive umber (b > 0) ad ot equal to 1 (b 1). The umber, N, must also be positive (N > 0) For the equatio below, chagig the Logarithm form to Epoetial form will solve for the variable. (These were covered i the Defiitio of a Log topic) log = 10 = 10 = 100 Note: Whe solvig ay Logarithm equatios, you should perform a check to see that the aswer(s) obtaied were cosistet with the defiitio of a Log. Similar to this, the questio below is solved usig the defiitio of a Log ad kowledge of equatios cotaiig epoets. log 5 = 5 = = ± 5 Cetre for Teachig ad Learig Academic Practice Academic Skills Page 1 T E ctl@scu.edu.au W [last edited o 13 July 015] CRICOS Provider: 0141G

2 Cetre for Teachig ad Learig Check: As the base of a Log must be (b > 0) ad (b 1), the solutio =-5 is ot cosistet with the defiitio. The solutio to this questio is =5 This type of questio ca become more comple depedig o the umbers. I this eample, a calculator is required. You may eed to review Epoets Equatios with Epoets. log =.1 =.1.1 = = to 3 d.p. A third type is where the variable becomes the epoet. For eample: log4 5 = ca be rewritte usig the defiitio of a log to be 5= 4 which ca be solved as a epoetial equatio (see below). However, a much easier method of solutio is to use the Chage of Base Rule. log4 5 = log 5 l 5 = alteratively log 4 l 4 = to 3 d.p. The secod major idea is based o equivalece. Put simply: If... log b M = log N the M = N b Note: There must be a sigle Log term o each side of the equals sig. Eample: log + log 5 = log 0 log 5 = log 0 Usig Log Rule 1 (Product Rule) 5 = 0 = 4 Cetre for Teachig ad Learig Academic Practice Academic Skills Page T E ctl@scu.edu.au W [last edited o 13 July 015] CRICOS Provider: 0141G

3 Cetre for Teachig ad Learig Eample: l = l 36 l l 36 Usig Rule 3 (Power Rule) = = 36 = ± 6 Check: The logarithm ca oly be foud of positive values oly, so the solutio is =6. Some equatios will cotai a mi of Log terms ad umbers, there are two strategies to solve this; (i) Rearrage to get Log terms o oe side ad umber(s) o the other, or (ii) Replace the umber with a equivalet Log. For eample - 1 is the same as log4 4, log55, log1010 etc. Eample: Solve for : log 4( + 4) = log43 Usig strategy (i) log ( + 4) = log log ( + 4) log 3 = log4 = = = = 48 = 44 = or Usig strategy (ii) log ( + 4) = log log ( + 4) log 4 = log log ( + 4) log 4 = log log4 = log = = 48 = 44 = Mied eamples: (i) Solve for. log9 = 0.5 log = = = 9 = 3 Cetre for Teachig ad Learig Academic Practice Academic Skills Page 3 T E ctl@scu.edu.au W [last edited o 13 July 015] CRICOS Provider: 0141G

4 Cetre for Teachig ad Learig (ii) Epress y i terms of : log y = log + log8 log y = log + log 8 log y = log + log 8 log y = log 8 y = 8 (iii) Epress y i terms of : l y = 3 + 4l l y = 3 + 4l 4 l y l 3 = y l = 3 4 y 3 = e 4 y = e 3 4 y= e 3 4 (iv) Solve for : log6 + log 6( + 5) = log6 + log 6( + 5) = log 6( + 5 ) = + 5= = 0 ( 4)( + 9) = 0 = 4 or 9 Check: Because it is ot possible to have the Log of a egative umber, the solutio is = 4. Video Solvig Logarithmic Equatios Cetre for Teachig ad Learig Academic Practice Academic Skills Page 4 T E ctl@scu.edu.au W [last edited o 13 July 015] CRICOS Provider: 0141G

5 Cetre for Teachig ad Learig Activity 1. Solve the followig Logarithmic Equatios. (a) log10 = (b) log5 = (c) log4 = 0.5 (d) log7 = (e) log ( + 4) = 3 (f) log9 = (g) log = (h) log3 = (i) ( ) 3 log + = 3 (j) log 4 4 = (k) log 9 = 0.5 (l) log 64 = (m) log ( 1) = () log 3 = (o) log 10 = 3 (p) log4 64 (q) log 10 = 4 (r) l( + 4) e = 7. Solve for i the Logarithmic Equatios. (a) log = log 4 (b) l = l 36 (c) log( + 5) = log + log 6 (d) 3log = log 4 (e) log = log + log 3 (f) log3 log 3( + 1) = log3 5 (g) log4 + log46 = log41 (h) 3log3 + log 3( ) = log35 (i) log 8( + ) = log8 (j) log = 5 log ( + 4) (k) log + log 3 = log 5 (l) log log( 1) = log 4 (m) log( 3) = 3 () l(4 ) + l = l (o) log 4( + 4) 3 = log43 (p) log + 3log = log 3. Solve for y i terms of the other variables preset. (a) log y = log + log 4 (b) 1 1 log y = log 5 + log (c) log y = log 4 log (d) log y+ log = log 4 + (e) log5 + log5 y = log (f) l y = l e 3l (g) log y+ log = 5 log 4 (h) log7 y = log7 5 + log7 + (i) 3l y = + 3l (j) (log4 y 3log 4 ) = 3 (k) y = e y + 1 (l) 10 = 3 = Cetre for Teachig ad Learig Academic Practice Academic Skills Page 5 T E ctl@scu.edu.au W [last edited o 13 July 015] CRICOS Provider: 0141G

6 Cetre for Teachig ad Learig Epoetial Equatios A equatio where the epoet (ide) is a variable or cotais a variable is called a epoetial equatio. A eample is 3 = 0. Let s cosider the eample below: If a ivestor deposits $0 000 i a accout that compouds at 5% p.a., how log is it before this amout has icreased to $30 000? (Compouded aually) Usig the compoud iterest formula gives: A= P(1 + i) = 0000(1.05) The first step is to rearrage the equatio to get the form a= b or b = a = = 1.05 The take the log of both sides. Because calculatios will eed to be performed, ordiary logarithms or atural logarithms should be used. 1.5 = 1.05 log1.5 = log1.05 takig the log of both sides log1.5 = log1.05 usig the third log law log1.5 = rearragig log1.05 = = 1.05 l1.5 = l1.05 takig the log of both sides l1.5 = l1.05 usig the third log law l1.5 = rearragig l1.05 = 8.31 This meas that it will take 8.31years (9 years i practice) for the growth i value to occur. This ca be checked by calculatig: This eample cotais a simple power =1.5 3 = 10 log 3 = log10 log 3 = log10 log10 = log 3 1 = log 3 =.096 Cetre for Teachig ad Learig Academic Practice Academic Skills Page 6 T E ctl@scu.edu.au W [last edited o 13 July 015] CRICOS Provider: 0141G

7 Cetre for Teachig ad Learig I this eample, the 4 must be rearraged to obtai the power as the subject before takig the logarithm of both sides. 4 7 = = 4 7 = 5 log 7 = log 5 log 7 = log 5 log 5 = log 7 = I this eample, there is a power as the subject; however, the epoet is more comple tha previous eamples = log1.5 = log ( 1) log1.5 = log log = log1.5 1 =.479 = = Eample: Joa wats to retire whe her superauatio fud reaches $ She ivests $1500 a moth (after ta) ito her superauatio fud. She assumes that the superauatio fud will retur 6%pa or 0.5%pm. How log before her goal is reached? This is a Future Value of a Auity, with r = 0.005, R = 1500 ad S = ( r) 1+ 1 S = R r ( ) = = 1500 ( ) 1 mult both sides by = divide both sides by = add 1 to both sides Cetre for Teachig ad Learig Academic Practice Academic Skills Page 7 T E ctl@scu.edu.au W [last edited o 13 July 015] CRICOS Provider: 0141G

8 Cetre for Teachig ad Learig To solve this epoetial equatio, logs are required = log.6666 = log1.005 take Log of both sides log.6666 = log1.005 use Log Law 3 log.6666 = log1.005 = 197 (rouded up) It will be 197 moths or 16 years 5 moths before Joa has eough moey. Video Solvig Epoetial Equatios Activity 1. Solve the followig epoetial equatios. (a) 4 = 0 (b) (c) = 0.15 (d) 4.6 = (e) 6 + = 40 (f) 11. = 0.6 (g) + 1 = 40 (h) = (i) 6 = 6 (j) 5 = =. If a ivestor deposits $ i a accout that compouds at 7% p.a., how log is it before this amout has icreased to $ ? (Compouded semi-aually). Use the compoud iterest formula: A= P(1 + i) Cetre for Teachig ad Learig Academic Practice Academic Skills Page 8 T E ctl@scu.edu.au W [last edited o 13 July 015] CRICOS Provider: 0141G