Exponential & Logarithmic Functions Chapter:___

Composite Functions Section:____

Composite Functions 1 Composite function: a function formed from two other functions 𝑓 and 𝑔 by substitution, which is then evaluated at 𝑥. 𝒇∘𝒈 𝒙 =𝒇( 𝒈(𝒙) ) 𝑓∘𝑔 reads “𝑓(𝑥) composed with 𝑔(𝑥)” . When you see 𝒇∘𝒈 , rewrite as 𝒇( 𝒈(𝒙) ) Domain: set of all domain values of g which make the output of g be the domain of f.

Composite Functions 2 Example: 𝑓 𝑥 = 𝑥 2 −3 and 𝑔 𝑥 =4 𝑥 2 Find: a) 𝑓∘𝑔 b) (𝑔∘𝑔)(1). a) 𝑓∘𝑔 Remember 𝑓( 𝑔(𝑥) ). In 𝑓∘𝑔, 𝑓 is written first. 𝑔(𝑥) goes into 𝑓(𝑥). Simplify. b) (𝑔∘𝑔)(1) Remember 𝑔( 𝑔 𝑥 ), then evaluate at 𝑥=1. In 𝑔∘𝑔, 𝑔 is written first. 𝑔(𝑥) goes into 𝑔(𝑥). Then simplify. Evaluate at 𝑥=1. 𝑓 𝑥 = 𝑥 2 −3 𝑓(𝟒 𝒙 𝟐 ) = (𝟒 𝒙 𝟐 ) 2 − 3 𝑓∘𝑔= 16𝑥 4 −3 𝑔 𝑥 = 4 𝑥 2 𝑔 𝟒 𝒙 𝟐 = 4(𝟒 𝒙 𝟐 ) 2 =4(16 𝑥 4 ) 𝑔∘𝑔= 64𝑥 4 𝑔∘𝑔 1 = 64 1 4 = 64

Composite Functions 3 Example: 𝑓 𝑥 = 3 𝑥−5 and 𝑔 𝑥 = 2 𝑥 Find domain of 𝑓∘𝑔. First find 𝑓∘𝑔. In 𝑓∘𝑔, 𝑓 is written first. 𝑔(𝑥) goes into 𝑓(𝑥). Then simplify. 𝑓 𝑥 = 3 𝑥−5 𝑓 𝟐 𝒙 = 3 𝟐 𝒙 −5 = 3∗(𝒙) 2 𝑥 ∗(𝒙)−5∗(𝒙) 𝑓∘𝑔= 3𝑥 2−5𝑥

Composite Functions 4 continued: Next, to find domain of 𝑓∘𝑔 : 𝑓∘𝑔= 3𝑥 2−5𝑥 continued: Next, to find domain of 𝑓∘𝑔 : First find restrictions of 𝑔 . Then find restrictions of 𝑓∘𝑔. Domain of 𝑓∘𝑔: 𝑔 𝑥 = 2 𝑥 𝑥≠0 𝑓∘𝑔= 3𝑥 2−5𝑥 Solve for 𝑥. 2−5𝑥≠0 −5𝑥≠−2 𝑥≠ 2 5 𝑥≠0, 𝑥≠ 2 5

Inverse Functions Section:____

Inverse Functions 1 One-to-One function: a function is one-to-one if any two different inputs in domain correspond to two different outputs in range. Set of ordered pairs: 𝑥’s and 𝑦’s cannot repeat. Graph: a horizontal line can intersect graph of a one-to-one function in at most one point. (Horizontal Line Test) Inverse function of 𝑓: a correspondence from the range of 𝑓 back to its domain, where 𝑓 is a one-to-one function Think of inverse functions as opposites in reverse order.

Inverse Functions 2 𝑓 must be one-to-one to have an inverse 𝑓 −1 . For inverses 𝑓 and 𝑓 −1 : Domain of 𝒇= range of 𝒇 −𝟏 , Range of 𝒇= domain of 𝒇 −𝟏 . Graphs of 𝑓 and 𝑓 −1 are symmetric about line 𝑦=𝑥. (Switch coordinates of 𝑓 to get graph of 𝑓 −1 .) To verify 𝑓 and 𝑓 −1 are inverses: show that 𝒇( 𝒇 −𝟏 (𝒙) )=𝒙 and 𝒇 −𝟏 ( 𝒇(𝒙) )=𝒙. To find inverse 𝑓 −1 from 𝑓: 1) Replace 𝒇(𝒙) with 𝒚. 2) Switch 𝒙 and 𝒚. 3) Solve for 𝒚.

Inverse Functions 3 Example: Find the inverse of the one-to-one function. Then state its domain and range. This is a one-to-one function, no 𝑥’s or 𝑦’s repeat. Switch domain & range to find inverse 𝑓 −1 . Domain of 𝑓 −1 : Range of 𝑓 −1 : { −2,5 , −1,3 , 3,7 , 4,12 , 0,−6 } { 5,−2 , 3,−1 , 7,3 , 12,4 , −6,0 } {5,3,7,12,−6} {−2,−1,3,4,0}

Inverse Functions 4 Example: Find the inverse of the one-to-one function: 1)Replace 𝑓(𝑥) with 𝑦. 2)Switch 𝑥 and 𝑦. 3)Solve for 𝑦. 𝑓 𝑥 = 9𝑥+1 𝑥+5 𝑦= 9𝑥+1 𝑥+5 𝑥= 9𝑦+1 𝑦+5 𝑥= 9𝑦+1 𝑦+5 (𝑦+5) (𝑦+5) Get y-terms together. 𝑥𝑦+5𝑥=9𝑦+1 𝑥𝑦 =9𝑦+1−5𝑥 Factor out common y. 𝑥𝑦−9𝑦= 1−5𝑥 𝑦(𝑥−9)=1−5𝑥 𝑦= 1−5𝑥 𝑥−9

Exponential Functions Section:____

Exponential Functions 1 Facts Laws of Exponents: Review p.420 Exponential Function: 𝒇 𝒙 = 𝒂 𝒙 Base 𝒂 is positive real number, but 𝒂≠𝟏. 𝒇 𝒙 = 𝒂 𝒙 is one-to-one function. If 𝒂>𝟏: 𝑓(𝑥) increases. OR If 𝟎<𝒂<𝟏: 𝑓(𝑥) decreases. Domain: (−∞,∞) Range: (0,∞) HA: 𝑦=0 x-int.: none y-int.: (0,1) Ratios of consecutive outputs =𝒂. Points on graph (𝑥 , 𝑦) −𝟏 , 𝟏 𝒂 (𝟎 , 𝟏) (𝟏 , 𝒂) ∞ −∞

Exponential Functions 2 Facts Definition: natural number 𝑒. Review p.427. To solve Exponential equations containing 𝒂 𝒙 , use the following property: If 𝒂 𝒖 = 𝒂 𝒗 , then 𝒖=𝒗 . Need to know Laws of Exponents(p.420) to rewrite equation when matching bases. Can solve with INTERSECT command on TI-83/84. Enter Y1 and Y2. Press 2nd TRACE . Select :intersect Exponential equations may have more than one solution.

Exponential Functions 3 Example: Solve 2 𝑥+1 =16. Type of equation: exponential. To solve, use property: If 𝒂 𝒖 = 𝒂 𝒗 , then 𝒖=𝒗 . Need 𝒂 𝒖 = 𝒂 𝒗 . Rewrite base 𝟏𝟔 to match base 𝟐. When bases are equal in 𝒂 𝒖 = 𝒂 𝒗 , solve 𝒖=𝒗. 2 𝑥+1 =𝟏𝟔 2 𝑥+1 = 2 4 Bases are equal. 𝑥+1=4 𝑥=3

Exponential Functions 4 Example: Solve 3 4 5𝑥+4 = 27 64 𝑥 . Type of equation: exponential. To solve, use property: If 𝒂 𝒖 = 𝒂 𝒗 , then 𝒖=𝒗 . Need 𝒂 𝒖 = 𝒂 𝒗 . Rewrite base 𝟐𝟕 𝟔𝟒 to match base 𝟑 𝟒 . When bases are equal in 𝒂 𝒖 = 𝒂 𝒗 , solve 𝒖=𝒗. 3 4 5𝑥+4 = 𝟐𝟕 𝟔𝟒 𝑥 3 4 5𝑥+4 = 𝟑 𝟒 𝟑 𝑥 Laws of Exponents (p.420) 𝑎 𝑠 𝑡 = 𝑎 𝑠∙𝑡 3 4 5𝑥+4 = 3 4 3𝑥 5𝑥+4=3𝑥 𝑥=−2

Exponential Functions 5 Example: Solve 3 𝑥 2 ∙ 27 𝑥 = 81 7 . Type of equation: exponential. To solve, use property: If 𝒂 𝒖 = 𝒂 𝒗 , then 𝒖=𝒗 . Need 𝒂 𝒖 = 𝒂 𝒗 . Rewrite base 𝟐𝟕 and 𝟖𝟏 to match base 𝟑. When bases are equal in 𝒂 𝒖 = 𝒂 𝒗 , solve 𝒖=𝒗. 3 𝑥 2 ∙ 𝟐𝟕 𝑥 = 𝟖𝟏 7 3 𝑥 2 ∙ 𝟑 𝟑 𝑥 = 𝟑 𝟒 7 Laws of Exponents (p.420) 𝑎 𝑠 𝑡 = 𝑎 𝑠∙𝑡 3 𝑥 2 ∙ 3 3𝑥 = 3 28 Laws of Exponents (p.420) 𝑎 𝑠 ∙ 𝑎 𝑡 = 𝑎 𝑠+𝑡 3 𝑥 2 +3𝑥 = 3 28 𝑥 2 +3𝑥=28 𝑥 2 +3𝑥−28=0 (𝑥+7)(𝑥−4)=0 𝑥+7=0 𝑥−4=0 𝑥= −7, 𝑥=4

Logarithmic Functions Section:____

Logarithmic Functions 1 Facts Logarithmic Function: 𝒇 𝒙 = 𝐥𝐨𝐠 𝒂 (𝒙) Base 𝒂 is positive real number, but 𝒂≠𝟏. 𝒇 𝒙 = 𝐥𝐨𝐠 𝒂 (𝒙) is a one-to-one function. 𝒇 𝒙 = 𝐥𝐨𝐠 𝒂 (𝒙) is inverse of 𝒇 𝒙 = 𝒂 𝒙 . If 𝒂>𝟏: 𝒇(𝒙) increases. OR If 𝟎<𝒂<𝟏: 𝒇(𝒙) decreases. Domain: (0,∞) Range: (−∞,∞) VA: 𝑥=0 x-int.: 1,0 y-int.: none Check solutions when solving Logarithmic equations. −∞ ∞ Points on graph (𝑥 , 𝑦) 𝟏 𝒂 ,−𝟏 (𝟏 , 𝟎) (𝒂 , 𝟏)

Logarithmic Functions 2 Facts 𝐥𝐨𝐠 𝒂 (𝒙) → “what power of 𝒂 gives me 𝒙?” Types of logarithms: Common Logarithm: 𝐥𝐨𝐠 (𝒙) → 𝐥𝐨𝐠 𝟏𝟎 (𝒙) Natural Logarithm: 𝐥𝐧 (𝒙) → 𝐥𝐨𝐠 𝒆 (𝒙) To rewrite Exponential & Logarithm equations, use the following property: OR 𝒙 =𝐥𝐨𝐠 𝒂 (𝒚) 𝒚 = 𝒂 𝒙 𝒙 = 𝒂 𝒚 𝒚 =𝐥𝐨𝐠 𝒂 (𝒙) becomes

Logarithmic Functions 3 Example: Find domain of 𝑓 𝑥 = log 9 𝑥+5 𝑥 . Type of function: logarithmic log 9 ( ) is defined only when output of 𝑥+5 𝑥 >0 . Find intervals where graph of 𝑌 1 = 𝑥+5 𝑥 is above x-axis. Domain: −∞,−5 𝑜𝑟 (0,∞) Remember HW 5.6(Rational Inequalities)? ∞ −∞ −𝟓 (

Logarithmic Functions 4 Example: Solve log 3 729 =5𝑥−4. Type of equation: To solve, use property: Rewrite log 3 (729) =5𝑥−4 so logarithm is on the right side. Solution set is { 2 } 𝒚 =𝐥𝐨𝐠 𝒂 (𝒙) logarithmic in this form 𝒙 = 𝒂 𝒚 becomes 5𝑥−4= 𝐥𝐨𝐠 𝟑 (729) Need 𝒂 𝒖 = 𝒂 𝒗 Rewrite 729 to match base 3. 729= 𝟑 5𝑥−4 𝟑 6 = 𝟑 5𝑥−4 Bases are equal, solve 𝒖=𝒗. 6=5𝑥−4 2=𝑥

Logarithmic Functions 5 Example: Solve 𝑒 2𝑥−5 =9. Type of equation: To solve, use property: Rewrite 𝑒 2𝑥−5 =9 so that exponential term is on the right side. Solution set is { ln 9 +5 2 } 𝒚 = 𝒂 𝒙 exponential in this form 𝒙 =𝐥𝐨𝐠 𝒂 (𝒚) becomes 9= 𝑒 2𝑥−5 2𝑥−5= log 𝑒 (9) Rewrite log 𝑒 ( ) →ln⁡( ) 2𝑥−5=ln⁡(9) Solve for 𝑥. 2𝑥= ln 9 +5 𝑥= ln 9 +5 2

Properties of Logarithms Section:____

Properties of Logarithms 1 Facts Since logarithms are exponents for some base 𝒂, logarithms follow basic properties of the Laws of Exponents(p.420). Properties of Logarithms will be used to expand or condense logarithmic expressions and equations. Properties of Logarithms. (𝑴, 𝑵, and 𝒂 are positive real numbers with 𝒂≠1, and 𝒑 is any real number.) 𝐥𝐨𝐠 𝒂 𝟏 =𝟎 𝐥𝐨𝐠 𝒂 𝒂 =𝟏 𝒂 𝐥𝐨𝐠 𝒂 (𝑴) =𝑴 𝐥𝐨𝐠 𝒂 ( 𝒂 𝒑 )=𝒑 When base 𝒂 and 𝐥𝐨𝐠 𝒂 ( ) act on one another as inverses, they cancel each other out.

Properties of Logarithms 2 Expand properties. Product Quotient Power Condense properties. 𝐥𝐨𝐠 𝒂 ( 𝑴∙𝑵 ) → 𝐥𝐨𝐠 𝒂 𝑴 + 𝐥𝐨𝐠 𝒂 𝑵 𝐥𝐨𝐠 𝒂 ( 𝑴 𝑵 ) → 𝐥𝐨𝐠 𝒂 𝑴 − 𝐥𝐨𝐠 𝒂 𝑵 𝐥𝐨𝐠 𝒂 ( 𝑴 𝒑 ) →𝒑∙ 𝐥𝐨𝐠 𝒂 𝑴 𝒑∙ 𝐥𝐨𝐠 𝒂 (𝑴) → 𝐥𝐨𝐠 𝒂 ( 𝑴 𝒑 ) 𝐥𝐨𝐠 𝒂 𝑴 + 𝐥𝐨𝐠 𝒂 𝑵 → 𝐥𝐨𝐠 𝒂 ( 𝑴∙𝑵 ) 𝐥𝐨𝐠 𝒂 𝑴 − 𝐥𝐨𝐠 𝒂 𝑵 → 𝐥𝐨𝐠 𝒂 ( 𝑴 𝑵 )

Properties of Logarithms 3 Equation properties. If 𝐥𝐨𝐠 𝒂 (𝑴) = 𝐥𝐨𝐠 𝒂 (𝑵) , then 𝑴=𝑵. If 𝑴=𝑵, then 𝐥𝐨𝐠 𝒂 (𝑴) = 𝒍𝒐𝒈 𝒂 (𝑵) . Change of Base formula.(Use to evaluate logarithms when base 𝒂 is not 10 or 𝑒.) 𝐥𝐨𝐠 𝒂 (𝑴) = 𝐥𝐨𝐠 (𝑴) 𝐥𝐨𝐠 (𝒂) or 𝐥𝐨𝐠 𝒂 (𝑴) = 𝐥𝐧 (𝑴) 𝐥𝐧 (𝒂)

Properties of Logarithms 4 Example: Use properties of logarithms to find exact value of log 2 5 ∙ log 5 (256) . Bases are different. Need Change of Base formula. Rewrite expression with formula. log 2 5 ∙ log 5 256 𝐥𝐨𝐠 𝒂 (𝑴) = 𝐥𝐨𝐠 (𝑴) 𝐥𝐨𝐠 (𝒂) 𝐥𝐨𝐠 (𝟓) 𝐥𝐨𝐠 (𝟐) ∙ 𝐥𝐨𝐠 (𝟐𝟓𝟔) 𝐥𝐨𝐠 (𝟓) = 𝐥𝐨𝐠 ( 𝟐 𝟖 ) 𝐥𝐨𝐠 (𝟐) = 𝐥𝐨𝐠 (𝟐𝟓𝟔) 𝐥𝐨𝐠 (𝟐) = 𝟖∙ 𝐥𝐨𝐠 (𝟐) 𝐥𝐨𝐠 (𝟐) = 𝟖

Properties of Logarithms 5 Example: Write the expression as a sum and/or difference of logarithms. 𝐥𝐨𝐠 𝒙(𝒙−𝟑) 𝒙+𝟖 𝟐 , 𝒙>𝟎 Need to expand. Use Expand properties. 𝐥𝐨𝐠 𝒙(𝒙−𝟑) 𝒙+𝟖 𝟐 Quotient Property 𝐥𝐨𝐠 𝒙(𝒙−𝟑) − 𝐥𝐨𝐠 (𝒙+𝟖) 𝟐 Product Property 𝐥𝐨𝐠 (𝒙) + 𝐥𝐨𝐠 (𝒙−𝟑) − 𝐥𝐨𝐠 (𝒙+𝟖) 𝟐 Power Property 𝐥𝐨𝐠 (𝒙) + 𝐥𝐨𝐠 (𝒙−𝟑) −𝟐 𝐥𝐨𝐠 (𝒙+𝟖) Answer: 𝐥𝐨𝐠 (𝒙) + 𝐥𝐨𝐠 (𝒙−𝟑) −𝟐 𝐥𝐨𝐠 (𝒙+𝟖)

Properties of Logarithms 6 Example: Find exact value of 𝒆 log 𝑒 3 (𝟔𝟒) Base is 𝑒. Need ln. Use Change of Base formula. 𝒆 log 𝒆 𝟑 (𝟔𝟒) = 𝒆 ln (𝟔𝟒) ln ( 𝒆 3 ) = 𝒆 ln (𝟔𝟒) 3 = 𝒆 [ 1 3 ln 𝟔𝟒 ] = 𝒆 [ ln ( 𝟔𝟒 𝟏 𝟑 ) ] = 𝒆 ln ( 3 𝟔𝟒 ) = 3 64 Answer: 𝟒

Solving Exponential & Logarithmic Equations Section:____

Solving Equations 1 Use all Properties to solve Equations. Equation solving checklist: Type of equation? Property needed? (TI-83/84 works too. Know it’s limitations.) Rewrite equation to model property, if necessary, then apply property to solve. Check all solutions of Logarithmic equations. (Solutions cannot make its input zero or negative.)

Properties of Logarithms 2 Example: Solve. 1 3 log 7 𝑥 =2 log 7 (4) Type of equation? Two Logarithms, same base. Use If 𝐥𝐨𝐠 𝒂 (𝑴) = 𝐥𝐨𝐠 𝒂 (𝑵) , then 𝑴=𝑵. Condense each logarithm, then solve. Logarithmic, remember 𝑥>0 1 3 log 7 𝑥 =2 log 7 (4) log 7 ( 𝑥 1 3 ) = log 7 ( 4 2 ) log 7 ( 3 𝑥 ) = log 7 (16) 3 𝑥 =16 𝑥= 16 3 𝑥=4096 Remember, 𝑥>0

Rewrite in Exponential form. Solving Equations 3 Example: Solve. 2 log 6 𝑥−5 + log 6 (4) =2 Type of equation? Condense logarithms, then rewrite in exponential form to solve. Logarithmic, 𝑥>5 2 log 6 (𝑥−5) + log 6 (4) =2 log 6 [ (𝑥−5) 2 ] + log 6 (4) =2 log 6 [ (𝑥−5) 2 ∙4] =2 Rewrite in Exponential form. 2= log 6 [ (𝑥−5) 2 ∙4] (𝑥−5) 2 ∙4= 6 2 (𝑥−5) 2 ∙4=36

Solving Equations 4 continued… (𝑥−5) 2 ∙4=36 (𝑥−5) 2 =9 (𝑥−5) 2 =± 9 (𝑥−5) 2 =± 9 𝑥−5=±3 𝑥−5=3 or 𝑥−5=−3 𝑥=8 or 𝑥=2 Remember, 𝑥>5.

Rewrite in Exponential form Solving Equations 5 Example: Solve. log (5𝑥+9) =1+ log (𝑥−1) Type of equation? Logarithmic, 𝑥>1 log (5𝑥+9) =1+ log 𝑥−1 Group logarithms log (5𝑥+9) − log 𝑥−1 =1 Quotient Property log 5𝑥+9 𝑥−1 =1 Rewrite in Exponential form 1= log 5𝑥+9 𝑥−1 5𝑥+9 𝑥−1 = 10 1 5𝑥+9 𝑥−1 =10

Solving Equations 6 continued… 5𝑥+9 𝑥−1 =10 5𝑥+9 =10𝑥−10 −5𝑥+9 =−10 Can solve with :intersect command on TI-83/84. Steps given in class. Be careful with decimal answers. (𝑥−1) 5𝑥+9 𝑥−1 =10(𝑥−1) 5𝑥+9 =10𝑥−10 −5𝑥+9 =−10 −5𝑥 =−19 𝑥 = 19 5

Solving Equations 7 Example: Solve. 2 1−3𝑥 = 7 𝑥 Type of equation? Bases can’t match, take ln( ) of both sides, then use properties to solve. Exponential ln ln⁡( ) 2 1−3𝑥 = 7 𝑥 Expand with Power Property [1−3𝑥] ln 2 =𝑥ln⁡(7) Use Distributive Property ln⁡(2)−3𝑥 ln 2 =𝑥ln⁡(7) Solve for x. ln 2 =𝑥 ln 7 +3𝑥 ln 2 ln 2 =𝑥[ ln 7 +3 ln 2 ] ln⁡(2) ln 7 +3ln⁡(2) =𝑥

Solving Equations 8 ( ) Example: Solve. 9 𝑥 + 3 1+𝑥 −10=0 Type of equation? Rewrite base 9 to match base 3, then solve. Exponential ( 3 2 ) 𝑥 + 3 1+𝑥 −10=0 ( 3 𝑥 ) 2 + 3 1 ∙ 3 𝑥 −10=0 ( 3 𝑥 ) 2 + 3 (3 𝑥 )−10=0 Factor as a Quadratic. ( ) =0 3 𝑥 + 5 3 𝑥 − 2 3 𝑥 +5=0 or 3 𝑥 −2=0 3 𝑥 is never negative. 3 𝑥 =−5 or 3 𝑥 =2 ln⁡(3 𝑥 )=ln⁡(2) 𝑥 ln⁡ 3 =ln⁡(2) 𝑥= ln⁡(2) ln⁡(3)

Finance Models Section:____

Finance Models 1 Formulas to solve Finance problems: The following are represented in given formulas: 𝑰 interest after 𝒕 years, 𝑨 amount after 𝒕 years, 𝑷 principal amount, 𝒓 interest rate, 𝒏 compounding periods per year. Simple Interest: Compound Interest: Continuous Compounding: Effective Rate of Interest: 𝑰=𝑷𝒓𝒕 𝑨=𝑷 𝟏+ 𝒓 𝒏 𝒏∙𝒕 𝑨=𝑷 𝒆 𝒓∙𝒕 Compounding 𝑛 times per year 𝒓 𝒆𝒇𝒇 = 𝟏+ 𝒓 𝒏 𝒏 −𝟏 Compounding continuously per year 𝒓 𝒆𝒇𝒇 = 𝒆 𝒓 −𝟏

Finance Models 2 Example: What amount results from investing $875 at 9.2% after 6 years if interest is compounded quarterly ? Use formula Example: Find effective rate of interest if 6.5% is compounded continuously ?(Round to 2 decimal places.) Use formula 𝒓 𝒆𝒇𝒇 = 𝒆 𝒓 −𝟏 𝑛=4 𝑨=𝑷 𝟏+ 𝒓 𝒏 𝒏∙𝒕 𝑨=(𝟖𝟕𝟓) 1+ (𝟎.𝟎𝟗𝟐) (𝟒) (𝟒 ∙ 𝟔) 𝑨=$ 𝟏𝟓𝟏𝟎.𝟏𝟔 𝒓 𝒆𝒇𝒇 = 𝒆 𝟎.𝟎𝟔𝟓 −𝟏 𝒓 𝒆𝒇𝒇 =𝟎.𝟎𝟔𝟕𝟏𝟓𝟗𝟎𝟐𝟒𝟒 % 𝟔.𝟕𝟐

Finance Models 3 Example: What interest rate compounded monthly will triple an investment in 8 years?(Round nearest tenth.) compounded monthly ? Find 𝒓. Use formula Investment tripled means 𝑨 is 𝟑𝑷. 𝑛=12 𝑨=𝑷 𝟏+ 𝒓 𝒏 𝒏∙𝒕 𝟑𝑷=𝑷 𝟏+ 𝒓 𝟏𝟐 𝟏𝟐∙𝟖 → 𝟑= 𝟏+ 𝒓 𝟏𝟐 𝟗𝟔 𝟑 𝟏 𝟗𝟔 = 𝟏+ 𝒓 𝟏𝟐 𝟏.𝟎𝟏𝟏𝟓𝟎𝟗𝟔𝟏 =𝟏+ 𝒓 𝟏𝟐 𝟎.𝟎𝟏𝟏𝟓𝟎𝟗𝟔𝟏 = 𝒓 𝟏𝟐 𝟎.𝟏𝟑𝟖𝟏𝟏𝟓𝟑𝟏𝟔𝟐=𝒓 % 𝟏𝟑.𝟖

Exponential Models 1 Formulas to solve Exponential Models. Growth & Decay: 𝑨(𝒕) final amount, 𝑨 𝟎 original amount, 𝒕 time 𝒌>0 (growth rate), 𝒌<0 (decay rate) Newton’s Law of Cooling: 𝒖(𝒕) final temperature of object, 𝒌 negative constant 𝒖 𝟎 initial temperature of object 𝑻 temperature of area where object is moved to Logistic Model: 𝑷(𝒕) final population, 𝒂 positive constant 𝒄 carrying capacity of population(growth), 𝒕 time, 𝒃>0 (growth rate), 𝒃<0 (decay rate) 𝑨(𝒕)= 𝑨 𝟎 𝒆 𝒌∙𝒕 𝒖 𝒕 =𝑻+( 𝒖 𝟎 −𝑻) 𝒆 𝒌∙𝒕 𝑷(𝒕)= 𝒄 𝟏+𝒂 𝒆 −(𝒃)∙𝒕

Exponential Models 2 Example: A radioactive material decays according to 𝑨 𝒕 =𝟕𝟎𝟎 𝒆 −𝟎.𝟎𝟑𝟖𝟏∙𝒕 . When 𝑡=0 days, how much radioactive material is still present? Find the decay rate. Graph 𝑨(𝒕) with window settings [0,1000,100]by[0,1, 0.25]. Press WINDOW . Enter each value given, then graph. At 𝑡=0, 𝐴 0 still present. 𝟕𝟎𝟎 grams −𝟑.𝟖𝟏 % A. B. C.

Exponential Models 3 continued. 𝑨 𝒕 =𝟕𝟎𝟎 𝒆 −𝟎.𝟎𝟑𝟖𝟏𝒕 . How much is left after t=5 years? (Round to nearest tenth.) Find the half-life of the radioactive material. (Round to nearest tenth of a year) 𝟓𝟕𝟖.𝟔 grams 𝑨(𝒕)=𝟕𝟎𝟎 𝒆 −𝟎.𝟎𝟑𝟖𝟏𝒕 𝟑𝟓𝟎=𝟕𝟎𝟎 𝒆 −𝟎.𝟎𝟑𝟖𝟏𝒕 Half of initial amount. 𝟏 𝟐 = 𝒆 −𝟎.𝟎𝟑𝟖𝟏𝒕 Solve for 𝑡. 𝐥𝐧 𝟏 𝟐 =−𝟎.𝟎𝟑𝟖𝟏𝒕 𝐥𝐧 𝟏 𝟐 −𝟎.𝟎𝟑𝟖𝟏 =𝒕 𝟏𝟖.𝟐 years

Need 𝒌 to form Cooling function. Exponential Models 4 𝒖 𝟎 Example: A thermometer initially reads 72°𝐹. At 1:00𝑝𝑚, it’s placed in a refrigerator having a constant temperature of 34°𝐹. After 2 minutes, the thermometer reads 58°𝐹. When will thermometer read 42°𝐹? 𝑻 𝒕 𝒖(𝒕) 𝒖 𝒕 =𝑻+( 𝒖 𝟎 −𝑻) 𝒆 𝒌∙𝒕 Need 𝒌 to form Cooling function. 58=34+(72−34) 𝑒 𝑘∙(2) ln (58−34) (72−34) ÷2=𝑘 −0.2297661647=𝑘 𝒖 𝒕 =𝟑𝟒+(𝟕𝟐−𝟑𝟒) 𝒆 −𝟎.𝟐𝟐𝟗𝟕𝟔𝟔𝟏𝟔𝟒𝟕𝒕 Cooling function

Properties of Logarithms 5 Example: If 𝑤= log 2 (45) and 𝑟= log 2 (9) , use properties of logarithms to find log 2 (5) . Bases are the same. Inputs 45 and 9 are related to input 5 by division. 5 becomes 45 9 , so use Quotient property. 𝐥𝐨𝐠 𝒂 ( 𝑴 𝑵 ) → 𝐥𝐨𝐠 𝒂 𝑴 − 𝐥𝐨𝐠 𝒂 𝑵 𝐥𝐨𝐠 𝟐 ( 𝟒𝟓 𝟗 ) → 𝐥𝐨𝐠 𝟐 (𝟓) → 𝐥𝐨𝐠 𝟐 𝟒𝟓 − 𝐥𝐨𝐠 𝟐 𝟗 𝐥𝐨𝐠 𝟐 (𝟓) = 𝒘−𝒓

Exponential Models 6 continued: When will thermometer read 42°𝐹? 𝒖 𝒕 =𝟑𝟒+(𝟕𝟐−𝟑𝟒) 𝒆 −𝟎.𝟐𝟐𝟗𝟕𝟔𝟔𝟏𝟔𝟒𝟕𝒕 Cooling function 𝟒𝟐=𝟑𝟒+(𝟕𝟐−𝟑𝟒) 𝒆 −𝟎.𝟐𝟐𝟗𝟕𝟔𝟔𝟏𝟔𝟒𝟕𝒕 ln (42−34) (72−34) ÷(−0.2297661647)=𝑡 𝟏:𝟎𝟒 𝟏:𝟎𝟕 𝟏:𝟎𝟗 𝟏:𝟏𝟐 PM. 6.781436336=𝑡

Exponential Models 7 continued: Graph 𝒖(𝒕) with window settings [0,20,2] by [0,100,10]. Use INTERSECT to find elapsed time for thermometer to read 35°𝐹. (Round to nearest tenth.) 𝒀 𝟏 =𝟑𝟒+(𝟕𝟐−𝟑𝟒) 𝒆 −𝟎.𝟐𝟐𝟗𝟕𝟔𝟔𝟏𝟔𝟒𝟕𝑿 𝒀 𝟐 =𝟑𝟓 1) Enter 2) Press 2nd TRACE . Select :intersect . 3) First curve? Move cursor near intersection. ENTER Second curve? ENTER Guess? Enter About 15.8 minutes need to elapse.