Exponential and Logarithmic Equations

Transcription

1 Sction. Eponntial and Logarithmic Equations *Thr ar basic stratgis for solving ponntial and arithmic quations. On is basd on th On-to-On Proprtis and th scond is basd on th Invrs Proprtis. On-to-On Proprtis 1. a = a y if and only if = y.. a = a y if and only if = y. Invrs Proprtis 1. a a =. a a = Stratgis for solving Eponntial and Logarithmic Equations: Rwrit th original quation in a form that allows th us of th On-to-On Proprtis of ponntial or arithmic functions to solv th quation. Rwrit an ponntial quation in arithmic form and apply th Invrs Proprty of arithmic functions to solv th quation. Rwrit a arithmic quation in ponntial form and apply th Invrs Proprty of ponntial functions to solv th quation. 1

2 Eampls using On-to-On Proprtis: Solv. CHAT Pr-Calculus Sction. (a) 6 6 (b) , 1 9 St th ponnts qual. 6 St th ponnts qual. (c) ln ln 0 (d) ln ln St argumnts qual. To Solv Using On-to-On Proprtis: 1. Isolat th variabl trm.. If it is ponntial, and th bass ar th sam, st th ponnts qual and solv.. If it is arithmic and th bass ar th sam, st th argumnts qual and solv.

3 Sction. Solving Eponntial Equations Using th Invrs Proprtis 1. Isolat th ponntial prssion.. Tak th arithm of ach sid.. Apply th Invrs Proprtis. *Not: Whn taking th of ach sid, us th sam bas as th bas on th ponntial prssion. Solving Logarithmic Equations Using th Invrs Proprtis 1. Isolat th arithmic prssion.. Writ ach sid in ponntial form.. Apply th Invrs Proprtis. *Not: Us th sam bas as th bas on th arithmic prssion. Eampls: Solv th following. (a) 11 (b) ln - Tak th of ach sid Us as bas on both sids. ln ln ln11 ln

4 Sction. (c) (d) () 16 (f) + 8 = 1 ln ln ln 0.69 ln ln ln

5 Sction. (g) 0 Thr ar ways to do this problm: 1. Tak th of ach sid: Uschang- of. - bas formula.. Tak th or ln of both sids and apply th Proprty of Eponnts or ln ln ln 0 ln ln 0 ln 0. **You can us ithr mthod.

6 Sction. (h) or ln 1 6 ln ln 6 ln ln 6 Us proprty of ponnts. ln (i) 0 0 Think of this as: ( ) ( ) 0 0 This is just a quadratic quation that can b factord. ( ( ln ) ( )( or ) 0 ln or ln ln ) 0 0 ln( ) *Not: ln(-) is impossibl so th only solution is ln. 6

7 Sction. (j) ln ln 66 ln 66 (ln 66) (k) (l) ln ln ln

8 Sction. (m) ln ln Th bass ar th sam, so st th argumnts qual. ( 0 )( 1) 0 or 1 Chck both in th original quation: Chck = Chck = -1 ln () ln ln ln ln ln() ln ( 1) ln( 1) You can stop hr, bcaus you cannot do ln(-1). So = -1 is an tranous solution. 8

9 Sction. (n) ( ) ( ) Us th Quotint Proprty for Logarithms. Solving an Equation on a Graphing Calculator Whn solving any quation, you can do it graphically by graphing th lft and right sids sparatly on your graphing calculator, and thn using th intrsct fatur to dtrmin th point of intrsction (i. th solution to th original quation.) 9

10 Sction. Eampl: Solv 0 Algbraically w gt: On your calculator, graph: y 1 = y =0 At th point of intrsction, y 1 = y, and so also =0. Th -coordinat of that point will b th solution to our quation. Adjust window to -10 < < 10 0 < y < Us [CALC] [intrsct] to find th point of intrsction. Th point of intrsction is (., 0). Thus, th solution is =.. 10

11 *In Gnral, to us th calculator to solv an quation graphically: CHAT Pr-Calculus Sction. 1. Graph th lft and right sids of th quation sparatly.. Us th [CALC] [intrsct] fatur to find th point of intrsction. Prss [CALC] [intrsct]. Us th lft and right arrows to mov th cursor along th first curv to gt as clos to th point of intrsction as possibl. Prss [ENTER]. Us th lft and right arrows to mov th cursor along th scond curv to gt as clos to th point of intrsction as possibl. Prss [ENTER] again. Prss [ENTER] to hav th calculator mak its guss. Th coordinats of th point of intrsction will appar at th bottom of th scrn. Not: It dos not mattr which quation you start with. Prssing th up and down arrows will mak th cursor jump from on graphd quations to th othr.. Th -coordinat of th point of intrsction will b th solution to th original quation. 11

12 Sction. Eampl: Find th solution using a graphing calculator. ( ) 1 Graph y 1 = (-) y = 1 Find th point of intrsction: Zoom in if ncssary. Us [CALC] [intrsct] to find th point of intrsction. Th point of intrsction is (., 1). Thus, th solution is.. Th solution w found arlir by doing this algbraically was = 10/.. 1

13 Sction. Applications: (1) How long would it tak for an invstmnt to doubl if th intrst wr compoundd continuously at 8%? Rmmbr th formula for continuously compounding intrst: A P If w want our principal to doubl, w must rplac A with P. W gt: P P rt 0.08t Now solv for t. Start by dividing both sids by P. ln t 0.08t ln 0.08t t 8.66 yars 1

14 Sction. () You hav $0,000 to invst. You nd to hav $0,000 to rtir in thirty yars. At what continuously compoundd intrst rat would you nd to invst to rach your goal? Solv for r. A P W nd 0,000 = 0,000 r(0) 0,000 7 ln 7 ln 7 r 0r ln 7 0 ln 0r 0r 0.06 rt 0,000 0r 6.% 1