CHAPTER 7: LOGARITHMIC and EXPONENTIAL FUNCTIONS SECTION 7.1: INVERSE FUNCTIONS. SECTION 7.2: ln x. . (Remember to simplify!)

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1 (Answrs to Erciss for Chaptr 7: Logarithmic and Eponntial Functions) A7 CHAPTER 7: LOGARITHMIC and EXPONENTIAL FUNCTIONS SECTION 7: INVERSE FUNCTIONS ) a) ; b) f ( ) = 4 4 ; c), which is th rciprocal of ) a) ; b) g ) = 5 a) 5 + b) c) d) ) / ; c), which is th rciprocal of SECTION 7: ln (Rmmbr to simplify!) t 40 Hint: Us th Powr Rul of Logarithms first 4t 7 + ( ln ) + ln Hint: Us th Powr Rul of Logarithms on th first trm f) w Hint: ln w g) = ln( w ) = ln w w w + ln w + ln w = ( + ln w) ( + ln w) ( + ln w) ( wln w + w) w ln w ) ( 4 + ) Hint: ln ( 4) 5 4 = ln( 4 + ) + ln 5ln( 4) by laws of logarithms

2 (Answrs to Erciss for Chaptr 7: Logarithmic and Eponntial Functions) A7 ) tan W thn (finally) hav: tan d = ln sc + C 4) a) and b) 7( sinθ )cos 6 θ Not: You may hav obtaind 7( tanθ )cos 7 θ for b) This is quivalnt to 7( sinθ )cos 6 θ, whr cosθ 0 Logarithmic Diffrntiation dos not apply for valus of θ that mak cosθ = 0 in this problm 5) 6) ( + ) a) ; b) + + ( + 5) + ; c) Your answr should ffctivly b th sam as your answr to part a) a) (, ) Hint: W rquir > 0 and ln > 0 b) ln + ln + ln c) ln ln d) On Dom f = (, ), f ( ) > 0 and f ( ) < 0 Thrfor, f is incrasing and th graph of f is concav down on th -intrval, ) Point-Slop Form: y ln = ( ), Slop-Intrcpt Form: y = + ln y = + ln 7) D ( ) =, and D ( ln ) = < whnvr > Not that > 0, and also > 0 whnvr > ; thrfor, and ln ar incrasing with rspct to on th intrval (, ) Altrnatly, bcaus D ( ln ) = > 0 whnvr >, w can conclud that th gap ln is incrasing on th intrval, incrasing fastr than ln is, and thrfor is 8) Hints: Lt y = n Apply Implicit Diffrntiation to both sids of ln y = ln( n )

3 (Answrs to Erciss for Chaptr 7: Logarithmic and Eponntial Functions) A7 SECTION 7: ) a) ) b) c) ( + ) 4 + Hint: ( + ) d) t +t Hint: t t simplifis to this ) + + ( + ) + f) ln + g) ln ln + h) cosθ sinθ i) sc tan j) 0 6r sc 4 6r ( )( ) simplifis to this ( + ) tan4 k) 6 csc cot l) θ cot( ) θ m) 4 4 cot( ) n) 0 a) y y 6 + b) ln ( 4 ) 6r ( )+ csc cot θ θ 4 csc ( ) 4 8( )cot y y y ( ) y + y 6 + csc ( ) ( 6 ) = ; c) ( + ) 6 ; 6 + 8

4 (Answrs to Erciss for Chaptr 7: Logarithmic and Eponntial Functions) A74 y y ) y ( sc y)tan y y y ( sc y)tan y y y tan y (bcaus sc y = y ; this is mor asily sn if w had takn th natural logarithm ( ln ) of both sids) 4) Hint: Th rat of chang of f with rspct to t is givn by f is givn by f ( t) Show that f t f ( t) In particular, f ( t) = b a bt 5) Dom( f ) = (, ) ( t) Th rat of dcay is qual to a positiv ral constant tims = b f t f ( ) > 0 for all ral valus of Obsrv that = f is vn, so its graph is symmtric about th y-ais HA: only y = 0, bcaus lim f = 0, and lim f f ( ) = π = 0 > 0 for all ral Obsrv that > 0 for all ral valus of CN: 0 Point at critical numbr: 0, f is incrasing on (, 0 f is dcrasing on 0, ), a local maimum point π f ( ) = ( π ) Obsrv that > 0 for all ral valus of PINs: and Concav up on (,, ) Concav down on, Both PINs corrspond to IPs:, π and, π

5 (Answrs to Erciss for Chaptr 7: Logarithmic and Eponntial Functions) A75 SECTION 74: INTEGRATION and LOG / EXP FUNCTIONS ) a) ln + C ln + C (Th Powr Rul for Logarithms can b usd to rprss som of th othr prssions in ths answrs) b) C C C C c) ln cos( ) + C C ln cos( ) ln sc( ) + C d) 5ln sin 5 + C ) 5 8 ln ( 4 + ) + C Not: 5 8 ln C = 5 8 ln ( 4 + ) + C, bcaus f) 4 + > 0 for all ral valus of C g) ln sc θ + h) ln i) + tan( θ + ) + C + C Not: ln C = ln ( 4 + 5) + C, += ( ) +> 0, bcaus = t + 6t + 9ln t + C t + t + 8ln t + C t + t + ln t 8 Hint: Epand th numrator by prforming th indicatd squar + C j) cot + 8ln csc cot C C cot + 8ln csc cot + 6, or C cot 8ln csc + cot + 6 k) cos ln l) π 7 + C C cos( ln ) + C m) ln + tan + C

6 (Answrs to Erciss for Chaptr 7: Logarithmic and Eponntial Functions) A76 n) + + C + + C + + C Hint: Epand th numrator by prforming th indicatd squar o) + C C p) ln + + C Not: ln + + C = ln( + ) + C, bcaus + > 0 for all ral valus of q) ln sc + tan( ) + C r) sin + C Hint: Us a Rciprocal Idntity + C s) ln cscθ cotθ + cosθ + C ln cscθ + cotθ + cosθ + C t) Hint: Us a Pythagoran ID ln + csc ) a) 0697, b) ln 069 ) (Lft to th radr) + C ln + csc 4) Hint: ln csc + cot = ln csc + cot 5) s( t) = t t + 7 6) π ( ) Hint: Stup is: + C = ln in ft s( t) = 44t m This is about 986 m csc + cot + 7 t t in ft csc cot csc cot π d from th Cylindr / Cylindrical Shll Mthod (6) Not : Obsrv that = 0 = π Not : π 7) π ln m Obsrv that both π ln this This is about 089 m > 0 for all ral valus of m and π ln m ar quivalnt to

7 (Answrs to Erciss for Chaptr 7: Logarithmic and Eponntial Functions) A77 ) SECTION 75: BEYOND NONNATURAL BASES a) + + ln 4 + ( ln ) ( + ) b) + ln + ln ln + ln c) ( )ln ( )ln0 d) π π + + π π lnπ π π ( π + lnπ ) sc 5t ) 5 ln f) ( ln6)r ln r sc( 5t ) sc 5t tan( 5t ) ( ln4) g) ln( + ) ( + )ln ( + ) + ( + ) h) ( sc )ln + tan tan, which can simplify to ( sc )ln + tan tan sc 5t, which can simplify to ) (Your answr should b quivalnt to th on for Ercis h) ) 4) a) ln 9π lnπ + C tan( 5t ) b) C 5ln7 ln7 + C = ln6,807 + C c) ( ln0)ln log + C ( ln0)ln ln + C (Ths ar quivalnt by th Chang-of-Bas Formula and th Quotint Rul for Logarithms Rmmbr that C can absorb constant trms) ln ln ln0 ln ln0 log by th Chang-of-Bas Formula