hw-5-sec-7.4-star-solns

hw-5-sec-7.4-star-solns - SECTlON 7.4* GENERAL LOGARITHMlC...

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Unformatted text preview: SECTlON 7.4* GENERAL LOGARITHMlC AND EXPONENTIAL FUNCTIONS 545 714* General Logarithmic and Exponential Functions (b) The domain of f (ac) = aI is R. (c) The range of f (03> = a1 [a ¢ 1] is (o, co). (ii) See Figure 3. (iii) See Figure 2. (d) (i) See Figure l. (b) The domain of fa) : loga x is (07 00). 2. (a) loga x is the number y such that ay : an. (c) The range of f(3:) : loga ac is R. (d) See Figure 9. : (€1n5)\/T : ex/Tln5 1012 Z (611110):z2 : 6.1221!) 10 @(Cosm)x : (elncosm)w : e 1. (a) log5 125 = 3 since 53 : 125. x1n(cosx)‘ $0053: : (elnz)cosz : e(ccxsar)(ln at) 1 _ _ 13_ 1 1 (b)10g3§7-— 3smce3 —33 27. 8. (a) log10 x/fi = log10 101/2 : g [analogous to (5) in Section 73*]. (b) 10g8 320 — log;8 5 :10g8 ig—O = log8 64 = 2 since 82 = 64. @(a) log2 6 ~10g2 15 + log2 20 z 105%) + 10g2 20 = 10g2(% ' 2O) : 10g2 8, and log2 8 = 3 since 23 : 8. [by Law 2] [by Law 1] (b) 10g3 100 a 10g3 18 — log3 50 = 10g3 (%) —10g3 50 : 10g3(1§)go) = log3(%), and logng) = #2 since 3—2 : %. 10. (a) log}; = #1 since a-1 = [0r: loga E : loga (f1 = #1] (logw 4+log10 7) : log104 _ 1 logm 7 Z 4 I 7 z (b) 10 10 o - (1°g10 4 +10%“) 7) # 10%10(4‘7) _— 10810 28 _ [On 10 10 10 28] 5 yzzo" y=51 yze" 12 11. All of these graphs approach 0 as w —> ~00, all of them pass through the point (0, 1), and all of them are increasing and approach 00 as m —* 00. The larger the base, the faster the function increases for m > 0, and the faster it approaches 0 as :E—v—oo. 12. The functions with bases greater than 1 (3m and 10$) are increasing, while those With bases less than 1 and (fi)? are decreasing. The graph of is the reflection of that 0f 31 about the y-axis, and the graph of is the reflection of that of 10”” about the y-axis. The graph of 107‘ increases more quickly than that of 3”” for :v > 0, and approaches 0 faster as x —+ —00. ":i SECTION 7.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS 20. We see from the graphs that for ac less than about 1.8, g(:c) intersect. Then f > 9(2) from a: z see that 9(2) > f In fact, 9 increases much more rapidly than f beyond that point. 32.5 *05 4375 50,000 1 5 ’g 5.2 0 7 21. lim (1.001)“° = 00 by Figure 1, since 1001 > 1. 23—400 lim as” = 0, so lirn (1.001)$ : 0. lid—‘00 22. By Figure 1, ifa > 1, :E—>*OO 23. lim 242 r lim 2“ [whereu = —t2] = 0 13—400 u‘afioo 33—3 —+0+. 11111 log 22—51%ka :lirn log 1:!00 10 10 z—>3+ t—>0+ 24. Lettza:2 —5ac+6.Asa:——>3+,t= (ac—2 [analogous to (4) in Section 72*]. C29 h(1t)=t3 — 3t :> h’(t) = 31:2 ~ 311113 = 33441 => g’(a:) = $44”: ln4 + 4x .4333 : m34z(acln4 + 4) 27. Using Formula 4 and the Chain Rule, y = 571/7” => y’ = 5’1/z(ln 5) [—1- (fix—2)] = 5‘1/I(ln 5)/av2 23. y = 10m”? 2 y’ = 10mn9(ln10)(sec2 9) 29_ I (211 +2-u)10 :> f/(u) : 10(2“ + 2‘“)9 i (2” + 2"”) = 10(2u + 2‘“)9 [2“ 1112 + 2'“ 1112 . (—1)] = 10 1112(2" + 2‘“)9(2“ — 2'”) 12 I2 d 2 r2 2 = 2 3 l : 3 __ w z 3 as y => y 2 (ln 2) dm (3 ) 2 (ln 2)3 (1n 3) (23:) , 1 d —3 3 t :10g2(1 _ 333) :> f z (1 r 3.10) ln2 E; (1 * 3m) 2 (1 — 3m) 1112 or (Bx — 1) ln2 _ ex(m+1) _x+1 1 (326 +6 ) (136m ln5 331115 x I 1 d I 32- f0?) =10g5($e) => f (9”) 2 3369211157115 356 )2 3312301115 6 function by first using logarithm properties. 1 1 ,_ 1 +_1_0r1+m e _a:ln5 1115 mln5 Another solution: We can change the form of th :1 m = x ’ = ' f(w) og5(me) 10g5m+10gse => “‘15) mln5+er1n5 547 : 53” > f(;r) 7— m5, and then near the point (1.8, 17.1) the curves 1.8 until at : 5. At (5, 3125) there is another point of intersection, and for a: > 5 we fi— SECTION 7.4‘ GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS C 549 005$ _ lnrcosx _e :> 44. f(m) = 2 m1) : elmm [111111— sm) + COWGN cos x , ‘ : m°°”[ —sm3:lnac] m This is reasonable, because the graph shows that f increases when f’ is positive. 210%“- 10t 2# 102 _ 101 #100—10# 90 g 1 ’ 111101’11110 11110” 11110 ‘11110 ' 1 47 103310 “3 dag : (1n m)/(1n 10) d3; : .— ln—SE dz. Now put u = In I, so du : —1— (110, and the expression becomes :3 x In 10 x a: 3 1 __ 1 l 2 __ 1 2 11110 “am—111ml?” +Cll‘2111100nm) +0 l 1 and we get / Cg? x dm = % ln10(log10 x)2 + C. Or: The substimtion u = 10g10 3” gives du : a: In 10 48 Lt # 2Thendufl2acda: so m2$2dx—l 2udu—l—2u—+C— 1 2I2+C -eu-$- — ’ ‘2 ‘21112 ‘21112 ‘ l . sin9 u 3“ 1 sinG 49. Letu=s1n0.Thenduzcos0d0and 3 costQ: 3 du:———+C=———3 +C. ; 1113 1113 1 i 2 dmz/lflzi1n1u|+0=fi1m2z+m+a Letu=2 +1.Thendu=2 ln2dm,so/2$+1 aha m2 1 l 1 2x 5x 0 51 2m 1 l 5‘ : I—x mflx Z —"—'_— —— ‘ 1A (2 5)d$+/0(5 2)” [1112 1115l21+lln5 1n2lo ‘ i 5 l l o ~1 _ _1___L _ 112-112 + i__2_ _ _1___1_ 5 ' 1112 1115 1n2 1115 1115 1112 1115 1112 _ 16 _ 1 § '51115 21112 52. Using disks, the Volume is V = fol 7r[10’“”]2 dx = 71' fol 10-21 dx. To evaluate the integral, we letu = —2m => ; du=—2dx,a:=0 é u=0,andm=1 :> u=——2,sowehave J 1 7r ‘2 7r 1 ’2 7r -2 997r I l Vzfl— 1"d =—— ———10“ :— 10 —1=_——— 1 1 2/0 0 “ 2l11110 l0 211110( ) 20011110 1 53. We see that the graphs of y = 2x and y = 1 + 3% intersect at a: z 0.6. We 15‘ f(w) = 2”” — 1 — 3—I and calculate f’(m) = 2a” ln2 + 3#z 1n 3, and using the formula 13n+1 = x" ~— f (5511) /f’(a:n) (Newton’s Method), we get $1 = 0-6, 1132 m 273 n 0600967. So, correct to six decimal places, the root m A 2 occurs at m = 0.600967. ...
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This note was uploaded on 10/10/2011 for the course M 408S taught by Professor Shirley during the Spring '11 term at UT Arlington.

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hw-5-sec-7.4-star-solns - SECTlON 7.4* GENERAL LOGARITHMlC...

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