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'()BCDEFTU]V^_mnz{!"#$|  0` r77f3/Ʊ` fff` KfxP` 7_/U<ff` HghXs3q̙` WXcklugti~^ӤO` ־3f3f` 33^` J%xiff3>?" dd@,?nFd@  d nF@ d`nF n?" dd@   @@``PP   @ ` ` p>>N0 rj(  6 T ~  "~  c B@CW DEF"d@ W @W @W W @`".~  c B,CW DEF"d@ W ,W ,@`". \ ~  "~  c B CDEF"@   @`"(4  c B C DEF"d@ @`"(W4  c BtC DEF"d@t t tt@`"~   c B CDEF"@ @`"([4W   c B CDEF"d@ @`"(4   c B CDEF"2@  @`"(4[   c BC DEF"d@ @`"_   c BC DEF"@ @`"  c BC DEF"@  @`"\  B* ?"0F  T Click to edit Master title style! !$  0- "0  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  04 "`P  B*   09 "`p   D*   0> "`0  D* H  0޽h ? fff80___PPT10. @F$ ShimmerQ 70N0 9(   T ~  "~~b ~ # "~  c B@CW DEF"d@ W @W @W W @`".~  c B,CW DEF"d@ W ,W ,@`".  c B CDEF"2@  @`"(4[  c BC DEF"@ @`"  c BC DEF"d@ @`"_b \~  # "\~   c B CDEF"@   @`"(4   c B C DEF"d@ @`"(W4   c BtC DEF"d@t t tt@`"~   c B CDEF"@ @`"([4W  c B CDEF"d@ @`"(4  c BC DEF"@  @`"\  H$  ?"p   T Click to edit Master title style! !  0  " `    W#Click to edit Master subtitle style$ $  0  "`P   B*   0Ȥ  "`@`    D*   0  "`0   D* H  0޽h ? fff80___PPT10. @F$)70 N0  (    0v`  ]%Logarithmic and Exponential Functions&( 2&P  0Lp OBy: Heather McGuire 3/21/08 H  0޽h ? 33___PPT10u..t6+D=' = @B +70N0    $ (    0@ X"The following is a review for students before they take the chapter test on exponential and logarithmic functions. They will be able to test their knowledge on the various topics and hopefully have feedback to help them study for the exam. Students will be able to work at their own pace.# 2#,  0U@ Cobb County Algebra 2 Standards M.ALGII.3.5 Logarithmic: Solve/Equations The learner will be able to determine value of common and the natural logarithms and antilogarithms using a calculator. Applies the change of base rule. M.ALGII.3.8 Logarithms: Definition/Properties The learner will be able to apply the definition and properties of logarithms to evaluate logarithms. Recognize and apply the inverse relationship of logarithms and exponential functions and graphs each function. M.ALGII.4.4 Predictions: Regression Techniques The learner will be able to solve exponential and logarithmic equations, such as those involving growth, decay, and compound interest. Make predictions from collected data by applying regression techniques. M.ALGII.4.8 Problem Solving: Relate The learner will be able to solve problems that relate concepts to other concepts and to practical applications using tools such as scientific or graphing calculators and computers.  ( 2 2 (-.#  0"`0@@AT0T0R  BZG%"`` JT0  3 A X n$0T0H  0޽h ? fff___PPT10.u`+{@{Dg' = @B D' = @BA?%,( < +O%,( < +D ' =%(%(D' =%(D=' =4@BBBB%(D' F=2y;BplayFrom(0.0)*<3<* DD'  =$ B B`B?<* %(/%,( < +/%,( < +/%,( < +)?+70N0    (    0$ `' R The number e* ( 2 00  0` LReview*( 2((`  c $A ??pP 3  09 The rules of exponents we have learned in the last 2 chapters also apply to e. Click the sound buttons to here the descriptions.6( 2L2`  c $A ?? 0  `   c $A  ??/ ^a   `   c $A b??# ; b  6"`@^T0T0R  BZG%"`` XT0"  0"`@`T0T0R  BG%"`` eT0F  S 0 F  S 0 ppn  3 A   , pT0n  3 A  P  , qT0n  3 A  P , rT0  3 A 0X s$0T0H  0޽h ?/@ fff___PPT10.wp!A+<D' = @B D' = @BA?%,( < +O%,( < +D ' =%(%(D' =%(D=' =4@BBBB%(D' =2y;BplayFrom(0.0)*<3<*D'  =0BcancelBubble@BA?%<(<*%<(<*_%(D' =%(D' =%(D=' =4@BBBB%(D' =2y;BplayFrom(0.0)*<3<*D:'  = B`B?<*%(/%,( < +/%,( < +/%,( < +)?D'  =0BcancelBubble@BA?%<(<*%<(<*_%(D' =%(D' =%(D=' =4@BBBB%(D'  =2y;BplayFrom(0.0)*<3<*D:'  = B`B?<*%(/%,( < +/%,( < +/%,( < +)?D'  =0BcancelBubble@BA?%<(<*%<(<*_%(D' =%(D' =%(D=' =4@BBBB%(D' @=2y;BplayFrom(0.0)*<3<*D:'  = B`B?<*%(/%,( < +/%,( < +/%,( < +)?DD'  =$ B B`B?<*%(/%,( < +/%,( < +/%,( < +)?+770N0 6.@ (  m  0l9`0 Now it is your turn to practice. Work these problems on a separate sheet of paper. Then click on the answer button to see the answer. Click the arrow at the bottom when you are ready to proceed to the next page. 2  6"`@T0T0R  BZG%"`0 T0`  c $A ??p  `   c $A !??pp  !   0s9` G 1T0 C Answer for a. 2   0x9 ` P  3T0 C Answer for b. 2`   c $A &?? p` &  0|9) ` `p 5T0 C Answer for c. 2H  0޽h ? fff___PPT10i.z@J+D=' = @B +p 70N0  u(    09P``  D Answer for a ( 2 (`  c $A '??p  '  c $A *??Z 8 *$D 0  0 9p  ,$0 UFirst you multiply the numbers. 2   09@p ` ,$0 lThen add the exponents of e.4 2   c $A -?? @ 8 -$D0R   0"` @mT0T0R   6"`P kT0   0X990  i5Click the u-turn arrow to return to the problem page.6 26   09 p 0 ,$ 0 QCongratulations you did it. 2H  0޽h ? fff___PPT10.~Yj+W2D' 9= @B D' = @BA?%,( < +O%,( < +D' =%(%(D ' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?bCB#ppt_x-.2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D' =-6Bwipe(right)*q3=-B%gradientSize: 0.1<*D' =%(DA' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-}6B!slide(fromLeft)*<3<*D ' =%( D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?bCB#ppt_x-.2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D' =-6Bwipe(right)*q3=-B%gradientSize: 0.1<*D"' =%(D' =4@BB%BB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* D' =+4 8?\CB#ppt_xBCB#ppt_xB*Y3>B ppt_x<* D-' =+4 8?hCB#ppt_y+1BCB#ppt_y-.03B*Y3>B ppt_y<* )?DI' d=+4 8?dCB#ppt_y-.03BCB#ppt_yB*Y3>B ppt_y<* %()?D' =%(pDP' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-6B#slide(fromRight)*<3<* ++0+0 ++0+0 ++0+ 0 +:70N0 91` (  R  0"` @T0T0R  6"`P T0  0990  i5Click the u-turn arrow to return to the problem page.6 26`  c $A ??P    0t9P``  D Answer for b ( 2 (   0$9p  SFirst you divide the numbers. 2`   c $A Z??P  Z   0D9p l  LThen subtract the exponents of e and place e where the largest exponent was.NM 2 !`  c $A \?? p \  09 p   QCongratulations you did it. 2H  0޽h ? fff___PPT10i.}C,+D=' = @B +70N0  (   R   0"` @T0T0R   6"`P T0   0990  i5Click the u-turn arrow to return to the problem page.6 26`   c $A ??@00    09P``  D Answer for c ( 2 (   09p l zDFirst you distribute the power to everything inside the parentheses.E 2E   0 9Jp P  q;Then use the rule of negative exponents to move everything.< 2<   09 p   QCongratulations you did it. 2`   c $A ^??E  ^`   c $A `?? @5 `H   0޽h ? fff___PPT10i.}C,+D=' = @B +70N0    6 (    0@9 `' P Logarithms* ( 2 00  09` LReview*( 2((  6"`@T0T0R  BZG%"`` T0"  0"`@T0T0R   BG%"`` T0c   09p`C ,$0 qThe only logarithms you can calculate on the calculator are base 10, the common log, and base e, the natural log.Br( 2^   :   09c ,$0 nLogarithms can be rewritten as exponential functions and exponential functions can be rewritten as Logarithms.o( 2o    09 0 } ,$0 logbaseanswer = exponent4 2         0|9 p 0,$0 baseexponent = answer 6( 2         Bp9#  ,$0 A translates to  H  0޽h ? fff___PPT10.pߵ+^ODF' = @B D' = @BA?%,( < +O%,( < +D8' =%(%(D' =%(D' =A@BBBB0B@?%(D' =1:Bvisible*o3>+B#style.visibility<* r%(D' =%(2DR' =A@BB BB0B@%(D' =-g6B fade*<3<* rD' =1:Bhidden*o3>+B#style.visibility<* r%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* D' =%(>D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* D)' =%(QFD' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* +P+0+ 0 ++0+ 0 ++0+ 0 ++0+ 0 ++0+ 0 ++0+0 + 70N0  (    6"`@T0T0R  BZG%"`` T0"  6"`@T0T0R  BG%"`` T0  0  RHow to evaluate logarithms( 2$   0$ `p  a+Rewrite the log as an exponential function., 2,0   B ' Blog5 25 5x = 25 x = 2`" 2    0 T ,$0 ;Some special rules for logarithms. If the base and the answer are the same, then the exponent is 1. log3 3 =1 ln e = 1~#( 2A 2( 2#D wH  0޽h ? fff___PPT10z.H +fD' = @B D' = @BA?%,( < +O%,( < +D' =%(%(D' =%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* +8+0+ 0 +t70N0  "(    0/  h,$0 BIf the base is raised to an exponent of log with the same base then the answer of the log is the solution. 6log6x = x eln 4 = 4z( 2l    p  0: ,$0 ^No matter what the base, if the answer is 1 the exponent is 0. log324 1 = 0 ln 1 = 0@?   B V  BB #k Z"More special rules for logarithms.# #   6"`@T0T0R  BZG%"`` T0"   6"`@T0T0R   BG%"`` T0H  0޽h ? fffrj___PPT10J.HЊ+D' = @B Di' = @BA?%,( < +O%,( < +D' =%(%(D' =%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-g6B fade*<3<*D' =%( D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-g6B fade*<3<*+p+0+0 ++0+0 +s170N0 PH   (      0   ?= x  2   0 9 `   s = 0 remember 250x = 1@ 2   0   t= 3 remember 10x = 1000@ 2   0ę ` 0   p= 1 remember 5x = 5@ 2   0ʙ P  p= 2 remember 2x = 4@ 2   0Й - v@Now it is your turn to practice& ! 2!    0ԙ @  ? Evaluate: 2  `   c $A g??X@' g`   c $A j??^ p j`   c $A q??p  q`   c $A t?? 6  t`   c $A w??K@kP w   0ٙ @ A Answer to b 2    0ȫ i @  A Answer to c 2    0@ 9 @  A Answer to d 2    0 @ A Answer to e 2    6"`@|T0T0R   BZG%"`` T0"   6"`@T0T0R   BG%"`` T0   0L P@ A Answer to a 2 H   0޽h ? fff""___PPT10".NP.iY+D/ '  = @B D'  =0BcancelBubble@BA?%<(<* %<(<* _%(DV' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<* D' =1:Bhidden*o3>+B#style.visibility<* %(Dh' =A@BB BB0B%()))D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* DR'  =0BcancelBubble@BA?%<(<* %<(<* _%(D&' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<* D' =1:Bhidden*o3>+B#style.visibility<* %(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* DR'  =0BcancelBubble@BA?%<(<* %<(<* _%(D&' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<* D' =1:Bhidden*o3>+B#style.visibility<* %(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* DR'  =0BcancelBubble@BA?%<(<* %<(<* _%(D&' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<* D' =1:Bhidden*o3>+B#style.visibility<* %(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* DR'  =0BcancelBubble@BA?%<(<* %<(<* _%(D&' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<* D' =1:Bhidden*o3>+B#style.visibility<* %(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* +0+0+ 0 ++0+ 0 ++0+ 0 ++0+ 0 ++0+ 0 ++0+ 0 ++0+ 0 ++0+ 0 ++0+ 0 ++0+ 0 +70N0  K(    6"`@T0T0R  BZG%"`` T0"  6"`@T0T0R  BG%"`` T0K  0P pj  YFor more help with evaluating logarithms visit http://themathpage.com/alg/logarithms.htmZ( 2Z("0"T00YH  0޽h ? fff___PPT10i.N X+D=' = @B + 70N0 @ $ (  $ $ 0X ` o)Expanding and Condensing Logarithms**( 2)00 $ 0( 00 fProperties of Logarithms2( 2$$ $ 6"`@T0T0R  $ BZG%"`` T0"  $ 6"`@T0T0R  $ BG%"`` T0  $ 0 00 `  \Product Property0 2$$`  $ c $A ??    $ 0 ' l 0  ]Quotient Property0 2$$` $ c $A ??    $ 0, l 00  ZPower Property0 2$$` $ c $A ??a = H $ 0޽h ? fff___PPT10i.N@$sZ+D=' = @B + 70N0   `((  ( ( 6"`@T0T0R  ( BZG%"`` T0"  ( 6"`@T0T0R  ( BG%"`` T0  ( 05 0P@ dTo Expand we use the properties of logarithms in the following order division multiplication powerse( 2e`  ( c $A ??e   ( 0`; @`  Z$First we apply the quotient property% 2%` ( c $A ??    ( 0? p p  W!Now we apply the product property" 2"` ( c $A ?? ^   ( 0C  @ V Last we apply the power property! 2!` ( c $A ??   ( 0H @  c-You have successfully expanded the logarithm.. 2.H ( 0޽h ? fff___PPT10i.N`\+D=' = @B ++70N0 p,+(  , , 6"`@T0T0R , BZG%"`` T0" , 6"`@T0T0R , BG%"`` T0 , 0O  `(Your turn to try Expand these logarithms)( 2)`  , c $A ??l `  , c $A ??  `  , c $A ??   , 0U ) @p A Answer to c 2 ` , c $A ??P{  , 0Y @0 A Answer to a 2 ` , c $A ??pP;    , 0\^ @  A Answer to b 2 ` , c $A  ??    H , 0޽h ? fff ___PPT10.N`\+D$' _ = @B DE'  =0BcancelBubble@BA?%<(<*,%<(<*,_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<*,D' =1:Bhidden*o3>+B#style.visibility<*,%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*,%(D' =-g6B fade*<3<*,DE'  =0BcancelBubble@BA?%<(<* ,%<(<* ,_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<* ,D' =1:Bhidden*o3>+B#style.visibility<* ,%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*,%(D' =-g6B fade*<3<*,DE'  =0BcancelBubble@BA?%<(<*,%<(<*,_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<*,D' =1:Bhidden*o3>+B#style.visibility<*,%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*,%(D' =-g6B fade*<3<*,++0+,0 ++0+,0 ++0+ ,0 + 70N0   0(  0 0 6"`@ T0T0R 0 BZG%"``  T0" 0 6"`@T0T0R 0 BG%"`` T0 0 0lq 0P@ fTo Condense we use the properties of logarithms in the following order powers division multiplicationg( 2g`  0 c $A ??c `    0 0w    W!First we apply the power property" 2"`  0 c $A! ??H - +    0 0D{  P   X"Now we apply the quotient property# 2#`  0 c $A" ??   0 0  P   X"Next we apply the product property# 2#` 0 c $A# ??P P   0 0̃ @P F e-You have successfully condensed you logarithm.( 2.H 0 0޽h ? fff___PPT10i.N`\+D=' = @B +-70N0 4-(  4 4 6"`@T0T0R 4 BZG%"`` T0" 4 6"`@T0T0R 4 BG%"`` T0 4 0D P b*Your turn to try Condense these logarithms+( 2+`  4 c $A( %??Ml %`  4 c $A$ -??p9  -`  4 c $A& 7?? x 7  4 08 ) p A Answer to c 2 `  4 c $A% )??kR ) 4 0, P0 A Answer to a 2 ` 4 c $A' /??p D;  / 4 0   A Answer to b 2 ` 4 c $A* 9??  9H 4 0޽h ? fff ___PPT10.N`\+D$'  = @B DE'  =0BcancelBubble@BA?%<(<*4%<(<*4_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<*4D' =1:Bhidden*o3>+B#style.visibility<*4%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* 4%(D' =-g6B fade*<3<* 4DE'  =0BcancelBubble@BA?%<(<*4%<(<*4_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<*4D' =1:Bhidden*o3>+B#style.visibility<*4%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*4%(D' =-g6B fade*<3<*4DE'  =0BcancelBubble@BA?%<(<* 4%<(<* 4_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<* 4D' =1:Bhidden*o3>+B#style.visibility<* 4%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*4%(D' =-g6B fade*<3<*4++0+ 40 ++0+40 ++0+40 +70N0 8P(  8 8 6"`@T0T0R 8 BZG%"`` T0" 8 6"`@T0T0R 8 BG%"`` T0P 8 04 pj  ^For more help with evaluating logarithms visit http://themathpage.com/alg/logarithms.htm#laws_( 2_("0'T00^H 8 0޽h ? fff___PPT10i.N X+D=' = @B +%70N0 }u < (  < < 6"`@T0T0R < BZG%"`` T0" < 6"`@T0T0R < BG%"`` T0 < 0 ` TChange of Base*( 2002  < 0 0 Up until now, the only logarithms we can use a calculator to calculate are base 10, the common log, and base e, the natural log.6( 2m  < 0  VThe Change of Base Formula ( 2 `  < c $A+ ??0 0F    < 0   YWith this formula we can use a calculator no matter what the original base of the log is.Z( 2ZH < 0޽h ? fff80___PPT10.770N0 >6@(  @ @ 6"`@T0T0R @ BZG%"`` T0" @ 6"`@T0T0R @ BG%"`` T0 @ 0 ` hChange of Base Example6( 20$$`  @ c $A) ??@P   @ 0 p V Apply the change of base formula! 2!`  @ c $A, ??   `  @ c $A- ??P <    @ 0< 0 p  8or 2(` @ c $A. ??G ;  ` @ c $A/ ??P C 2  @ 0 p P  8or 2(H @ 0޽h ? fff80___PPT10.c70N0 XP D(  D`  D c $A7 ??&   `  D c $A3 ??   D 0d4 0 XYour turn to try Apply the change of base and approximate the answer to 3 decimal placesY( 2Y` D c $A0 ??i ` D c $A2 ?? nG   D 09 7 D Answer to a. ( 2   D 0H> p   D Answer to b. ( 2   D 6"`@T0T0R  D BZG%"`` T0" D 6"`@T0T0R D BG%"`` T0H D 0޽h ? fff  ___PPT10s .p+D ' E = @B DE'  =0BcancelBubble@BA?%<(<*D%<(<*D_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<*DD' =1:Bhidden*o3>+B#style.visibility<*D%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* D%(D' =-g6B fade*<3<* DDE'  =0BcancelBubble@BA?%<(<* D%<(<* D_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<* DD' =1:Bhidden*o3>+B#style.visibility<* D%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* D%(D' =-g6B fade*<3<* D+p+0+D0 ++0+ D0 + 70N0   H (  H H 6"`@T0T0R H BZG%"`` T0" H 6"`@T0T0R H BG%"`` T0 H 0P  c+Finding the inverse functions of logarithms,( 2,(d  H 0U h  pThis process is the same as finding the inverse of any function: Step 1. Swap x and y Step 2. Solve for y.xA( 20 2Q  H 0\  `  OExample: find the inverse 2`  H c $A4 ??P |x    H 0h  P  v Swap x and yN 2`  H c $A1 ??0 |X  H 0g    b,Solve for y by rewriting in exponential form- 2-` H c $A5 ??( P  H 0\t P i3Congratulations you have found the inverse function4 24H H 0޽h ? fff___PPT10i.Ѝ9+D=' = @B +)70N0 0 L(  L` L c $A: ??/ `  L 6"`@T0T0R L BZG%"`` T0" L 6"`@T0T0R L BG%"`` T0 L 0 0 t<Your turn to try Find the inverse of the following functions=( 2=`  L c $A6 ??6M `  L c $A; ??    L 0\  `  z*Hint: when solving for y move the 3 first.4+ 2  L 0̒ 6 }  D Answer to b. ( 2 ` L c $A8 ??P    L 0  D Answer to a. ( 2 H L 0޽h ? fff  ___PPT10s .N++D '  = @B DE'  =0BcancelBubble@BA?%<(<* L%<(<* L_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<* LD' =1:Bhidden*o3>+B#style.visibility<* L%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*L%(D' =-g6B fade*<3<*LDE'  =0BcancelBubble@BA?%<(<* L%<(<* L_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<* LD' =1:Bhidden*o3>+B#style.visibility<* L%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*L%(D' =-g6B fade*<3<*L+p+0+ L0 ++0+ L0 + 70N0 @ Px(  P P 6"`@T0T0R P BZG%"`` T0" P 6"`@T0T0R P BG%"`` T0 P 0p pJ TExponential Growth and Decay( 2(  P 0h @@  @The function represents Exponential Growth if a > 0 and b > 1. 8A 2  P 0    {y = abx6( 2   P 0 @P  CThe function represents Exponential Decay if a > 0 and 0 < b < 1. 8D 2  P 0 a @  WExamples of Exponential Growth ( 2 `  P c $A< ?? `P  P 0 ` p   WExamples of Exponential Growth ( 2 H P 0޽h ? fff___PPT10i.`j++D=' = @B +P 70N0 O G ` X(  X X 6"`@T0T0R X BZG%"`` T0" X 6"`@T0T0R X BG%"`` T0 X 0 pJ TExponential Growth and Decay( 2( X 0 @  @The function represents Exponential Growth if a > 0 and b > 1. 8A 2 X 0  m {y = abx6( 2      X 0 @P  CThe function represents Exponential Decay if a > 0 and 0 < b < 1. 8D 2  X 0 a @  WExamples of Exponential Growth ( 2 `  X c $A< ?? `P   X 0h ` p   VExamples of Exponential Decay( 2`  X c $AN ?? TL H X 0޽h ? fff___PPT10i.`j++D=' = @B +% 70N0 $  P T(  T T 6"`@T0T0R T BZG%"`` T0" T 6"`@T0T0R T BG%"`` T0 T 0 pJ TExponential Growth and Decay( 2(  T 0  m {y = aex6( 2    ,  T 0 @ When working with e, the rules for exponential growth and decay change slightly. We can now just look to the exponent to determine growth or decay.( 2 T 0  &  SThe function represents Exponential Growth if a > 0 and the exponent is positive. 8T 2* T 0 P &  NThe function represents Exponential Decay if a > 0 and exponent is negative. 8O 2&` T c $A= ??0  ` T c $A> ??0 ` H T 0޽h ? fff___PPT10i.p++D=' = @B +70N0 o g p\ (  \ \ 0\4 0@6 n8This exponential decay because the is between 0 and 1.9 29 \ 6"`@T0T0R \ BZG%"`` T0" \ 6"`@T0T0R \ BG%"`` T0  \ 0: 0 ZYour turn to try Determine if the function is exponential growth or decay and explain why?[( 2[`  \ c $A@ ??L `  \ c $AA ?? p  `  \ c $A9 ?? t   \ 0?  @  D Answer to b. ( 2  \ 0pB  D Answer to a. ( 2  \ 0F ) p D Answer to c. ( 2  \ 0lJ  @  >This exponential growth because the e has a positive exponent.4? 2$ \ 0HO  @ =This exponential decay because the e has a negative exponent.D> #CCCH \ 0޽h ? fff___PPT10.PPg3+DK' [ = @B DR'  =0BcancelBubble@BA?%<(<*\%<(<*\_%(D&' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<*\D' =1:Bhidden*o3>+B#style.visibility<*\%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*\%(D' =-g6B fade*<3<*\DR'  =0BcancelBubble@BA?%<(<* \%<(<* \_%(D&' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<* \D' =1:Bhidden*o3>+B#style.visibility<* \%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*\%(D' =-g6B fade*<3<*\DR'  =0BcancelBubble@BA?%<(<*\%<(<*\_%(D&' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<*\D' =1:Bhidden*o3>+B#style.visibility<*\%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*\%(D' =-g6B fade*<3<*\+P+0+\0 ++0+ \0 ++0+\0 ++0+\0 ++0+\0 ++0+\0 +~ 70N0 } u `(  ` ` 6"`@T0T0R ` BZG%"`` T0" ` 6"`@T0T0R ` BG%"`` T0 ` 0l  e-Solving Exponential and Logarithmic Equations.( 2.(  ` 0Tr @` zBCan we rewrite the log as an exponential function to solve for x? C( 2C`  ` c $AC ??~ X  ` 0 @`  ` c $AB ?? G ` ` c $AD ??@ X ` 0p`@` ` c $A? ??p }  X ` 0 ` ` ` c $AE ??p   ` ` c $AF ??' pp X ` 0 ` F ` S   H ` 0޽h ? ` fff___PPT10i.P[|4+D=' = @B +> 70N0   d (  d d 6"`@T0T0R d BZG%"`` T0" d 6"`@T0T0R d BG%"`` T0 d 0̉  e-Solving Exponential and Logarithmic Equations.( 2.(  d 0 @F kDo we have the same base or can we rewrite the bases as common bases? If so then the exponents are equal. l( 2l`  d c $AI ??#  X  d 0`  d c $AM ??` !  x  d <AG ??P  X d 0@` d c $AH ??   X d 0 P ` d c $AK ??k 6 x d <AL ??P X d 00@x d <AJ ??_ = X d 00 X d 000x d <AP ?? `X F d S f  H d 0޽h ? d fff80___PPT10.W: 70N0 q i h (  h h 6"`@T0T0R h BZG%"`` T0" h 6"`@T0T0R h BG%"`` T0 h 0  e-Solving Exponential and Logarithmic Equations.( 2.(  h 0 @ We can NOT rewrite both sides with the same base. Now we must take the log of both sides using the base that has the exponent.( 2` h c $AO  ??c @   X h 0 p ` h c $AQ ??`   X h 0 P ` h c $AR ?? 7   h 0 A a  :Now use the change of base to give an approximation for x.6;( 28` h c $AR ??  X h 0@P` h c $AS ??  H h 0޽h ? fff80___PPT10.W; 70N0 K C l (  l l 6"`@DT0T0R l BZG%"`` >T0" l 6"`@T0T0R l BG%"`` T0 l 0؇ ^&Your turn to try Solve these equations'( 2'`  l c $AT ??P p `  l c $AV ?? @@ `  l c $AU ??< '   `  l c $AW ?? P  ` l c $AX ??z   l 0\u%@ 4T0 A Answer to b 2  l 0 i%i @  6T0 A Answer to c 2  l 0@V%9 @  8T0 A Answer to d 2  l 0%@ :T0 A Answer to e 2  l 0tP@ T0 A Answer to a 2 H l 0޽h ? fff+#___PPT10.[0s+D' K9= @B D'  =0BcancelBubble@BA?%<(<*l%<(<*l_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<*lD' =1:Bhidden*o3>+B#style.visibility<*l%(D'  =0BcancelBubble@BA?%<(<*l%<(<*l_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<*lD' =1:Bhidden*o3>+B#style.visibility<*l%(D'  =0BcancelBubble@BA?%<(<*l%<(<*l_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<*lD' =1:Bhidden*o3>+B#style.visibility<*l%(D'  =0BcancelBubble@BA?%<(<*l%<(<*l_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<*lD' =1:Bhidden*o3>+B#style.visibility<*l%(D'  =0BcancelBubble@BA?%<(<*l%<(<*l_%(D' =%(D' =%(D6' =A@BB BB0B%(D' =-g6B fade*<3<*lD' =1:Bhidden*o3>+B#style.visibility<*l%(++0+l0 ++0+l0 ++0+l0 ++0+l0 ++0+l0 +70N0 VN p(  pR p 6"` @T0T0R p <"`P T0 p 0 P``  D Answer for a ( 2 (` p c $AT ??P p   p 0\;   &,$0 v@The bases are the same so set the exponents equal to each other.A 2A  p c $AY  ??YO@8  $D 0  p 04NP Pp,$0 FNow solve for x. 2  p c $AZ #?? 2PG 8 #$D 0  p c $A\ &?? P 8 &$D 0  p c $A[ )?? 4P 8 )$D 0H p 0޽h ? fff___PPT10.[bs+F5DD2' = @B D' = @BA?%,( < +O%,( < +D$' =%(%(D' =%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*p%(D' =-g6B fade*<3<*pD' =%(DA' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* p%(D' =-}6B!slide(fromLeft)*<3<* pD' =%( D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* p%(D' =-g6B fade*<3<* pD' =%( DA' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* p%(D' =-}6B!slide(fromLeft)*<3<* pD' =%(|DA' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* p%(D' =-}6B!slide(fromLeft)*<3<* pD' =%(LDA' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* p%(D' =-}6B!slide(fromLeft)*<3<* p+p+0+p0 ++0+ p0 +&70N0   x(  xR x 6"` @T0T0R x <"`P T0 x 0 P``  D Answer for b ( 2 ( x 0pD ,$0 RRewrite in exponential form. 2 x c $A] 0??B8 0$D 0 x 0WDP Pp,$0 >Simplify 2  x c $A^ 2??) @Dp 8 2$D 0`  x c $AV .??% .H x 0޽h ? fff  ___PPT10 .[bs+YPxD ' = @B D ' = @BA?%,( < +O%,( < +D ' =%(%(D' =%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*x%(D' =-g6B fade*<3<*xD' =%(DA' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*x%(D' =-}6B!slide(fromLeft)*<3<*xD' =%( D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*x%(D' =-g6B fade*<3<*xD' =%( DA' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*x%(D' =-}6B!slide(fromLeft)*<3<*x+p+0+x0 ++0+x0 +70N0 LD0 |(  |R | 6"` @T0T0R | <"`P T0 | 0 x P``  D Answer for c ( 2 ( | 0tN ,$0 `Rewrite in exponential form.( C | c $Aa J??B8 J$D 0 | 0 P Pp,$0 RNow solve for x.& C  | c $A_ L??9 M 8 L$D 0  | c $Ab N?? `` 8 N$D 0`  | c $A` H??P' $ H  | c $Ac Q?? 0@8 Q$D 0H | 0޽h ? fff___PPT10.[bs+'D>' = @B D' = @BA?%,( < +O%,( < +D0' =%(%(D' =%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*|%(D' =-g6B fade*<3<*|D' =%(DA' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*|%(D' =-}6B!slide(fromLeft)*<3<*|D' =%( D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*|%(D' =-g6B fade*<3<*|D' =%(DE' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* |%(D' =-6B%slide(fromBottom)*<3<* |D' =%(XDE' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* |%(D' =-6B%slide(fromBottom)*<3<* |D' =%((#DE' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* |%(D' =-6B%slide(fromBottom)*<3<* |+p+0+|0 ++0+|0 +!70N0  y @  (  R  6"` @T0T0R  <"`P T0  0( P``  D Answer for d ( 2 (  0/X  ,$0 DFirst we must rewrite the bases as the same base raised to exponent.(E DC  c $Ad ]??s 8 ]$D 0  0hHX P p,$0 i'Use your rules of exponents to simplify&( 'C  c $Ae _??.  8 _$D 0   c $Af a?? #@ 8 a$D 0   c $Ak c?? 008 c$D  0`   c $AW Z??Pb Z   0 uX ` f ,$0 p.Bases are the same so the exponents are equal.&/ .C   0lX   ,$0 N Solve for x.&  CH  0޽h ? fff___PPT10z.[bs+hDn' = @B D)' = @BA?%,( < +O%,( < +D`' =%(%(D' =%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-g6B fade*<3<*D' =%(DA' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-}6B!slide(fromLeft)*<3<*D' =%( D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-g6B fade*<3<*D' =%(DE' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B%slide(fromBottom)*<3<*D' =%(XD8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* D' =%(@DE' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-6B%slide(fromBottom)*<3<* D' =%('D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* D' =%(*DE' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-6B%slide(fromBottom)*<3<* ++0+0 ++0+0 ++0+ 0 ++0+ 0 +70N0 ZRP (  R  6"` @T0T0R  <"`P T0  0 P``  D Answer for e ( 2 (8  0XX  ,$0 `We can NOT rewrite both sides to have the same base, so we must take the same log of both sides.(a `C  c $Ag e??8 e$D 0  0HN  ,$0 >Now use the properties of Logarithms to simplify the left side(? >C  c $Ai g?? > 8 g$D 0   c $Ah i??  8 i$D 0`   c $AX [??4& [&   0X * 0 ,$0 0Use the change of base formula to approximate x.F1 .CCCH  0޽h ? fff-%___PPT10.[bs+D1' = @B D' = @BA?%,( < +O%,( < +D#' =%(%(D' =%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-g6B fade*<3<*D' =%(DA' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-}6B!slide(fromLeft)*<3<*D' =%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-g6B fade*<3<*D' =%(|DE' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B%slide(fromBottom)*<3<*D' =%(LD8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* D' =%(4!DE' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-6B%slide(fromBottom)*<3<* ++0+0 ++0+0 ++0+ 0 +70N0 t((  t t 0tiX Z  XCongratulations you have reached the end of the review. 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