Math 153: Lecture Notes For Chapter 5

Transcription

1 Mth 5: Lecture Notes For Chpter 5 Section 5.: Eponentil Function f()= Emple : grph f ) = ( if = f() Emple : Grph ) f ( ) = b) g ( ) = c) h ( ) = ( ) f() g() h() ) b) c) - -

2 Emple : Use the grph of = to grph Shifting ) = + b) = Shifting c) = + d) = Reflecting e) = f) + = ) Shift up b) Shift down c) Shift left d) Shift right e) Reflect in the -is, then down f) Reflect in the -is, then up Eponentil function where > 0: Emples: f ( ) = f ( ) = > < Eponentil Growth = = = Eponentil Dec

3 Emple : Solve the following equtions: ) = b) 5 = 5 d) = 6 e) 7 = + 9 c) f) 8 = 9 + = 7 ( ) Compound Interest: A = P + r n nt P: the principl, mount invested A: the new blnce t: the time in ers r: the rte, (in deciml form) n: the number of times it is compounded. Emple 5: Suppose tht $5000 is deposited in sving ccount t the rte of 6% per er. Find the totl mount on deposit t the end of ers if the interest is: P = $5000, r = 6%, t = ers ) compounded nnull, n = : A = 5000( /) ()() = 5000(.06) () = $6.8 b) compounded seminnull, n =: A = 5000( /) ()() = 5000(.0) (8) = $6.85 c) compounded qurterl, n = : A = 5000( /) ()() = 5000(.05) (6) = $6.9 d) compounded monthl, n =: A = 5000( /) ()() = 5000(.005) (8) = $65. e) compounded dil, n =65: A = 5000( /65) (65)() = 5000(.0006) (60) = $656. r + n Note: For Compounded Annull where n =, A = P = P( + r) t The following emple is from the book: 6. Find n eponentil function of the form f ( ) = b tht hs the given -intercept nd psses through the point P. -intercept 6; P(, /) 8. Find n eponentil function of the form f ( ) = b c tht hs the given horizontl smptote nd - + intercept nd psses through the point P. = 7; -intercept 5; P(, 8.5) nt

4 Section 5.: Nturl Eponentil Function f()=e Compound Interest: A nt r = P + (from section 5.) n Continuous Compound Interest: Continuous compounding mens compound ever instnt, consider investment of $ for er t 00% interest rte. the following tble shows the compound interest tht results s the number of compounding periods increses: P = $; r = 00% = ; t = er Compounded Number of periods per er Compound Amount nnull (+/) = $ monthl (+/) = $.60 dil 60 (+/60) 60 = $.75 hourl 860 (+/860) 860 = $.78 ech minute 58,00 (+/58,00) 58,00 = $.787 As the tble shows, s n increses in size, the limiting vlue of P is the specil number e =.788. If the interest is compounded continuousl for t ers t rte of r per er, then the compounded mount is given b: A = P.e rt Emple : Suppose tht $5000 is deposited in sving ccount t the rte of 6% per er. Find the totl mount on deposit t the end of ers if the interest is compounded continuousl. Solution: P =$5000, r = 6%, t = ers A= 5000.e (0.06)() = 5000.(.75) = $656. Emple : Find the mount to be invested t rte of 8% compounded continuousl in order to get $,90 in 6 ers. Solution: A = $,90, r = 0.08, t = 6 ers. $,90 = P.e (0.08)(6), then P = $8000 Emple : The price of certin computer is given s: 0.0t P( t) = P o. e where P o is the initil price. Approimte the percentge of the originl price fter 5 ers.

5 Emple : grph f ( ) =, f ( ) = e ) f ( ) = b) f ( ) = e 0 b) ) Emple 5: grph f ( ) = e nd f ( ) = e ) f ( ) = e b) f ( ) = e 0 - ) b) The grph of Eponentil Function f ) = ( nd the Nturl Eponentil Function f ( ) = e : Approches but does not cross the -is. The -is is Horizontl Asmptote. nd e cnnot be = 0 nd does not cross the -is (where = 0). Emple 6: Find the zeros of f: ) f ( ) = e + e b) f ( ) = 6 e e c) f e ( ) = + 8 e Emple 7: Simplif the epression: ( e e ) ( e + e ) ( e + e ) 5

6 Section 5.: Logrithm Function If f ) = f ( where > 0, find the inverse ( ) Solution: We cn use the procedure of section.8 Rewrite it s =. Replce b nd b : = Isolte to find f ( ) : =...? To isolte when > 0, > 0 nd, we use logrithm: If = = log Eponentil Form then Logrithmic Form is the bse in both forms Emple. Convert the following from eponentil form to logrithmic form: ) = Answer: log = b) = 8 Answer: log 8 = c) 9 / = Answer: log 9 = / d) 0 = 00 Answer: log 0 00 = Emple. Convert the following from logrithmic to eponentil form: ) log 9 = Answer: = 9 b) log 5 5 = Answer: 5 = 5 c) = log Answer: = d) log = Answer: = Emple. Find the number if possible: ) log b) log 5 5 c) = log log d) log 5 (-) > 0 (no logrithm for zero or negtive) > 0, 6

7 Emple.. Solve for : ) log = 5 Answer: = 5 = b) log 5 = 0 Answer: = c) log = - Answer: = /8 d) log (+) = Answer: = 7 Emple 5: Grph the following if = : ) = ; or = b) = log ; or = log then = = = ) Emple 6: Use the grph of = log to grph : Shifting ) = (log ) + b) = (log ) ) Shift up b) Shift down c) Shift left d) Shift right - Shifting c) = log ( + ) d) = log ( ) Reflecting e) = log or = - f) = log ( ) or e) Reflect in the -is f) Reflect in the -is = 7

8 Common nd Nturl Logrithms: = log The number is clled the logrithmic bse If = 0, then it is log 0 or log nd it is clled Common Logrithm (vilble in clcultor s log) If = e, then it is log e or ln nd it is clled Nturl Logrithm (vilble in clcultor s ln) log 0 = log log = ln Common Logrithm e Nturl Logrithm Emple 7. Convert the following from eponentil form to logrithmic form: ) 0 = b) e = + c) e = 0. 5 e d) = 5 Emple 8. Convert the following from logrithmic to eponentil form: ) log = b) ln( + ) = c) ln = + d) ln = 5 I. Solve for ) log 9 = Answer: = 9 b) log = Answer: = c) ln = Answer: = e (wh?) log = ; log e e = or ln e = II. Solve for ) log 9 = 0 Answer: = b) log = 0 Answer: = c) ln = 0 Answer: = (wh?) log = 0 ; log e = 0 or ln = 0 8

9 From section.8, If f () nd f ( ) re inverse to ech others, then: f ( f ( )) = ; f ( f ( )) = III. If f ( ) = then f ( ) = log, nd: f ( f ( )) = f ( ) = log = IV. If f ( ) = then f ( ) = log, nd: f ( f ( )) = f (log ) = log = Summr: log ln I. log = ln e = II. log = 0 ln = 0 III. log = ln e = IV. log = ln e = Emple 9. Find the number: ) c) e) log 5 0 b) log 0 ln e d) / ln e / ln e Emple 0. Solve the eqution: ) log ( + ) = log ( ) b) log ( ) = ln c) log = d) e = 6 ln e) e = 0. 5 Note: It is esier to solve the stor problems of this section (problems: 5, 59, 6, 69, 70, nd 7) fter covering section

10 Section 5.: Properties of Logrithms Rule Formul Emples Wrning I) Multipliction Rule log ( u. w) = log u + log w ln( u. w) = ln u + ln w log 5 = log 5 + log ln u.ln w ln u + ln w ln( u + w) ln u + ln w II) Division Rule III) Power Rule log ln u w = log u w u log = ln u ln w c log u = c log u ln = cln u u c w 5 ln u log = log 5 log ln u ln w ln w log 5 = log 5 (ln u ) c cln u Emple. Epress in term of logrithms: ) log b) log c) log z 5 z w Emple. Epress s one logrithm: ) log log + log z b) log log log z c) log( 9) log( + ) log Emple. Solve for : ) log + log( ) = b) log + log ( ) = c) log ( ) log ( ) = d) ln( + ) + ln( ) = 0 e) ln( ) = f) ln = + ln( + ) g) log( + ) log = log h) log + log = log 5 + log ( ) i) log + log( ) = log j) ln( ) ln( + ) = log k) log ( + ) = log ( ) + log 9 + Emple. Sketch the grph of f: ) f ( ) = log (6 ) Solution: f ( ) = log (6 ) = log (. ) = log + log. = + log. Shift the grph = log up units b) f ( ) = log ( ) Solution: f ( ) = log ( ) = log ( ). Verticll stretched the grph = log b fctor of. 0

11 Section 5.5: Eponentil nd Logrithmic Equtions. Isolte the eponentil epression on one side of the eqution. Tke the logrithm of ech side nd use the power rule of logrithm to bring down the eponent. Solve for the vrible Section 5. Rules: Section 5. Rules: log = ln e = log ( u. w) = log u + log w log = 0 ln = 0 log = ln e = log u w = log u log w log = ln e = c log u = c log u Emple. Solve for : ) = 7 c) 8. e = 0 d) 5 + b) = 7 = + e) + = f) log( + ) = + log( ) g) log( 0) log( ) = log h) 6( ) = 6 i) ( ) + ( ) = 9 Chnging Logrithm Bse: log log u b = log Emple. Find the following: (round the nswer to deciml plces) ) log 5 8 b) log 7 c) log u b Emple. Solve for in two different ws : = 7 ) b tking the log of both sides b) b chnging the log bse

12 Emple. Solve for without clcultor: ) log(log ) = b) log 5 = c) e + e = 5 Emple 5. Use common logrithms to solve for in term of : 0 = 0 Continuous Growth nd Dec (see section 5.) Growth: rt A = P. e Dec: A = P. e rt Doubling Time = ln r Hlf-life = ln r A: the new blnce, popultion.. P: the principl, mount invested, initil mount.. t: the time in ers r: the continuous rte, (in deciml form) Emple 6. If $5000 invested t rte of 9% per er compounded continuousl, how long does it tke to double? Emple 7. How much mone must be deposited tod to mount to $000 in 0 ers t 5% compounded continuousl? Emple 8. If the popultion in certin countr is 6. millions nd growing t continuous rte of.% per er. When will the popultion rech millions? Emple 9. In 998, reserchers found n ivor tusk tht hd lost 8% of its crbon-. How old ws the tusk 0000t if the dec ws given s P( t) = P o e. Emple 0. A fisher stocks pond with 000 oung trout. The number of the originl trout still live fter t 0.t ers is given b: P( t) = 000e ) How mn trout left fter 6 months b) At wht time will there be 00 of the originl trout left? Emple. The mss m(t) remining fter t ers from smple of rdioctive Cesium-7 is given b 0.00t m( t) = m0e, where m 0 is the initil mss of the substnce. Find the hlf-life of Cesium-7. (Definition of hlf-life: The hlf-life substnce is the time it tkes for one-hlf the initil mount in given smple to dec.) Emple.. The hlf-life of certin rdioctive substnce is ds. If there re 0 grms initill: ) find the rte. b) when will the substnce be reduced to grms?