Lecture 10, 11, and 12 Notes
Lecture 10, 11, and 12 Notes Math 240
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This 15 page Class Notes was uploaded by AnnMarie on Friday October 9, 2015. The Class Notes belongs to Math 240 at Louisiana Tech University taught by Jonathan B Walters in Fall 2015. Since its upload, it has received 47 views. For similar materials see Precalculus in Mathematics (M) at Louisiana Tech University.
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Date Created: 10/09/15
Chapter 41 Exponential Functions Definition The exponential function base bis defined for all real numbers x by fx bx where b gt O and Z9750 Graphs of Exponential Functions The exponential function fx bquot bgt0b1 has a domain of R and range 0 00 The line y O the xaxis is a horizontal asymptote of f The graph of f has one of the following shapes fxaxf0ragt1 fxaxf0r0ltalt1 Note Both of these graphs are onetoone The most commonly used exponential Functions are 1 fx ex 2 fx 10x 3 fx 2x Shifting stretching re ecting properties still apply when performing transformations to exponential function Example Sketch fx 2x 1 1 Solution Shift the graph of fx 2quot to the right 1 unit and then shift upward 1 unit 114 Properties of Exponential Functions 1 b0 1 2 9119 3 bxbybxy 4 g lax y 5 196Vbe 6 axbx 61179C 7 3 x Example Write the expression as a single power of the form bx 42x3 X 8xl 8xx44x Solution First change 4 into 22 and 8 into 23 222x3 X 23x 1 X Second using the 5th property of exponential functions to get 243C6 X 233C3 233C X 283C Third using the 3rd property of exponential functions to get 24x63x3 23x8x Fourth combining like terms to get 27X3 21 13C Fifth using the 4th property of exponential functions to get 27X311X Finally combine like terms again to get 23 4X Compound Interest Let A be the amount accrued after t years Interest is calculated on the principle P and interest earned before new compound period The equation for compound interest is as follows AU Plt1 ifquot where At amount after t years P principle r interest rate per year 72 number of times interest is compounded per year t number of years Example J eaanIarc wants to buy a car in 5 years His car will cost 32000 He found a great stock that guarantees 9 return per year compounded quarterly How much should he deposit today to get 32000 in 5 years Solution What we know At 32000 r 009 n 4 t 5 Substituting values for At r n andt 32 000 P1 02 20 Divide both sides by 1 to get 32000 P 10 Solve for P P 5392020532 20 50613 Chapter 42 Natural Exponential Functions Definition The natural exponential function is the exponential function f x ex with base e It is often referred to as the exponential function The graph of fx ex appears as Shifting re ecting stretching and shrinking still apply Example Graph fx ex Solution When learning about trigonometric hyperbolic functions you will the natural 3 k exponent in both the numerator and divisor The hyperbolic cosine function is E quot coshx WTW The graph of the hyperbolic cosine looks E 11 E like the graph to the right The hyperbolic sine function looks like Sinhx gage x The graph of the hyperbolic sine looks like the graph to the left 43 Logarithmic Functions Recall That fx bx is a onetoone function The inverse function of fx bx is the logarithmic function f1x 0gbx Definition Let b be a positive number with b 79 1 The logarithmic function with base I denoted by 0gb is defined by logbxy gt byx So 0gbx is the exponent to which the base I must be raised to give x Granhs of LOgarithmic Functions M W fx10gbx b gt 1 fx10gbx 0ltblt1 an f1x bx 3 fx 0gbx where b gt 1 f1x bx fX 10gbx where O lt b lt 1 The domain and range off 10 is as follows Df 000 Rf 007 00 43 Continued logbx y ltgt by x Recall for inverses ff1x xandf1fx x blogbx x wherex gt O logbbx x for all x Example Evaluate the following a log464 b log32i7 C 510g5x2 2 Solution a log464 log443 3 3 b 0832L7 0836 3 C 5log5x22 x22 Commonly Used Logarithms 1 Natural Log logex lnx 2 Common Log log10x logx 3 Dyadic Log log2x lgx Properties of Logarithms 1 logb1 O 2 logbb l 3 logbbx x 4 logbuv Zogbulogbv 5 logbe logbw 10310 6 logbuv vlogbu Example Find the domain and sketch lnx 2 Solution x 2 gt O x gt 2 Domain 200 5 7 I I 11 Example Combine the following into a single algorithm a 310g4 0g21 20g8 0gl4 b 3lns 5140 4lnt2 1 Solution a zog43 0g21 log82 0gl4 log64 log21 log64logl4 log21 logl4 10g 10g b lns3 moi lnt2 14 MSW lnt2 14 magi1t Change of Base Formula Let ab gt 0 but 6119791 then 0gax Example Evaluate log aT Solution ZOgVEGT 111 Example Write 0gJC2T25W in the form A logx2 11 B logx2 11 C 0gx3 2 Solution 10gx2 11 Iogw 11gtx3 2f 45 Logarithmic and Exponential Equations Note bx by gt x y since fx b2 is onetoone Example Solve 24quot2 4 C2 0 Solution 24x 2 4x2 2 0 24x 2 22x2 0 24x 2 22x2 2 0 24x 2 22xx 4x 2 2x2 2x2 4x 2 O 2x2 2x 1 O 2x2 x x 1 O 2x2 x x 1 0 2xx 1 1x 1 O 2x 12 0 x 1 What if we can t get the same base Example Solve for x a 3quot2 7 b 8e2x 20 Solution a ln3x2 1147 Take the logarithm of both sides x 2ln3 1147 Bring down x 2 x 2 2 Divide n3 on both sides x 222 2 Subtract 2 from both sides b e2x 28 0 Divide both sides by 8 1462 0 16 Take the logarithm of both sides 2xlne ln Bring down 2x 2x mg ne1 2 Divide both sides by 2 Exponential Equations that look like quadratic equations can be solved using u substitution Example Solve for x ezx ex 6 0 Solution Letting u equot uz u 6 O u22u 3u 6 0 Since 2 gtlt 3 6 expand the equation u22u 3u60 uu2 3u20 u2u 3O u 2 u 3 716quot ln3 Take the logarithm of both sides xlne ln3 Bring down the x x 113 Since In e 1 Example Solve for x x22x 2x18 0 Solution 2xx2 18 0 Since 2quot is never zero divide both sides by 2quot x2 18 Subtract 18 from both sides x i V Square both sides Logarithmic Equations 10gbx laggm x y Example Solve for x lnx 3 lnx 5 ln2x 7 Solution lnx 3x 5 ln2x 7 Multiply x3 and x5 according to Property Rule 4 of Log lnx2 8x 15 ln2x 7 x28x152x7 x28x 2x15 7O x26x80 x24x2x80 x24x2x80 xx42x40 x2x40 x 2 x 4 Check for extraneous solutions x 2 gt ln1ln3 713 Oln3 ln3 x 4 gt ln 1 ln1 ln 1 ln1 is not possible Example Solve 0g6x 6 log6x 3 2 Solution log6x2 3x 6x 18 2 x2 3x 18 26 Rewrite 2 into the logarithmic exponential form x23x 18 36O x2 3x 54 O x29x 6x 54 0 x2 9x 6x 54 O xx9 6x90 x 6x90 x 6 x 9 Check for extraneous solutions x 6 gt log612 log63 2 log636 2 log662 2 x 9 gt log6 3log6 12 2 Unable to evaluate Example Solve for x a 2amp2 b 0g2log3x 4 Solution a2 2 4Zog5x 1 210537500 21057500 5 12 36 75 b log2log3x 4 24 log3x Example Solve for x 2quot 102x 3 0 Solution Let u 2x u 10u13 0 u 30 uu 1 3Ou u2 103u0 u23u 10 O u25u 2u 10O u25u 2u 10 O uu5 2u50 u 2u50 u 2 u 5 It is not possible for u 5 to be a solution 2x2 x1