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# MAT 170 Study Guide MAT 170

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This 41 page Study Guide was uploaded by India Scott on Thursday October 1, 2015. The Study Guide belongs to MAT 170 at Arizona State University taught by Fida Sleiman in Summer 2015. Since its upload, it has received 92 views. For similar materials see Precalculus in Liberal Arts at Arizona State University.

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Date Created: 10/01/15

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395 9 m 3 E E g g 5 3 m ag w wmwmwxawxww g 2 W awquot W 7 W E a zzwawmmm awm mwm w gm aa w W n A Q m x A gw ag a 5 Wm 95 a y Q a J 2 guy 4 W17 W am 395 E 5 33 E w 2 Ex My k 3 3 E a 12 bewr w w W 1 5 H y A m x mm Ea Em gm 3 Wz w mm M wm 2 a quota t W 1 3394 3 V w amm ww 2 m ow M f wg 3 a 3 3 g w u 5 a a a me mmwm Q3 mm m a aw my 9 V emu a a g t 4 E gaff a WK W3 A g H u 3 a a N g a k a W w W A A pa 5 mg WWA g ew I x V a WM 5 kg m wwwma x m 55a 2W 1 a y W 6232 M a w m NF tax a a WW gaw 939 3 r g i x b 2 2 K a 1 E W x 2 g i imm mm amp mwmg mamx 39 mm V w I mnmwwmmvywymaw wmm 39rmw Kssz wwm 5 g 7 wwvw 39 gamma W NM a A w g 2 g m a y W M gm 33 A1 U amp b a M x m A k 3 ma a A i Kw w wm Mag my m m a V J n b 3 mm m 5 5 3 a w mam mp z w 1ng gm m3 M w m m a h w 4 gm M w r a a m Kawa E w a m 33 was w s gmwww w TE g a qg g w g 3 Section34 Solving Exponential and Logarithmic Equations I Exponential Equation is an equation containing a variable in an exponent 39 1 Method 1 Solving exponential equations by expressing each side as a power of the same base If W 9N then M 2N Solve a 53 6 125 rewrite both sides with same base 5 53x6 53 So 3x 6 3 and 3x 9 and so x 3 We say the solution set is 3 b 8 2 2 4quotquot3 rewrite both sides with same base 2 23x2 22x3 23x6 22x 6 So 3x6 2x6 and x 12 so the solution set is 12 2 Method 2 Solving exponential equations by taking the natural or common logarithm of both sides after isolating the exponential expression 39 Solve a 4x 15 we can t write both sides with the same base so take log of both sides log 4x 2 long use the power rule to get xlog4 log 15 divide both sides by log 4 to get x long 195 solution set is 10gls d log4 log4 Or take In of both sides and x iii 11215 z 195 I b 7e 5 58 since the given base is e take 1n not log of both sides 7e 2 58 5 so 7e 2 63 and ezquot 9 divided both sides by 7 Take In of both sides to get Inc 3 ln9 and 2x 1n 9 the inverse functions In and e undo each other ln9 Solving for x we get x 2 z 110 c 5x2 42x3 lnsxm 1n42x3 x 2 ln5 2x3 ln4 39xln5 21n5 2xln4 31n4 xlnS 2xln4 21n5 31n4 xln5 21n4 21n5 31n4 21n53ln4 z634 ln5 21n4 The solution set is W 1n5 21n4 d e 8ex 7 0 Think of ean y so e yzand we get ex lequot 7O y28y70so y l y 70 ex10 01 ex 7O 6quot 1 Or ex 7 lnex ln1 Or lnex ln7 lenl orxln7 So x O or x ln7 and the solution set is 0 ln7 II Logarithmic Equation equation containing a variable in a logarithmic expression Method 1 Solving logarithmic equations by rewriting the equation in exponential form If log b M C then bC M Include in the solutions only the values that make M gt 0 since the domain of a logarithmic function is the set of positive real numbers Solve a log2x 4 x 3 23 z x 4 8x4andx 12 Before concluding that the solution set is 12 we have to check and make sure we are taking the logarithm of a positive real number since logarithms of nonpositive numbers are not de ned so replace x by 12 in the given logarithmic expression 12 4 is 8 which is positive and therefore it s possible to take its logarithm Now we conclude that the solution set is 12 b 4ln3x 8 ln3x 84 2 ln3x 2 and so 62 2 3x and x z 33 This value for x makes 3x positive 2 and so its logarithm is de ned The solution set is 53 r c log X logX3 1 use the product rule to rewrite it as a single logarithm log XX 3 1 logx2 3x130101x2 3x x2400 XS x2 0 sox S 00rx20and sox5 OFX32 Checking x5 makes both the x and the X 3 positive and so their logarithms are de ned so X 5 is a solution x x 2 makes us take the logarithm of a negative which is not de ned so x 39 w 2 is not a solution The solution set is5 Method 2 Solving logarithmic equations by using the following one to one property If logb M 2 10gb N then M N Include in the solution set only the values of X that make Mgt0 and Ngt0 Solve a lnX3 ln7X 23 lnx1 ln X 3 ln 7x 23 x 1 w o x 3 1 multiply both Sides by x1 x x1 X 3 7X 23 39 x22x3x7x23 x2 9x20 0 And X 5x 4 0 so x 5 or X 4 Checking Both the x4 and XZS make each of the X 3 7x 23 and the xl positive and so their logarithms are de ned The solution set is 4 5 Application The formula A 1599023St models the population of Florida A in millions t years after the year 2000 a What was the population of Florida in 2000 b B When will the population in Florida reach 192 millions a In 2000 t z 0 since it is 0 years after 2000 A x 159 90235 159 1 since e 1 159 millions b 192 159 9023 2 60235 take In of both SidBS 1n192 11190235 159 1111 9 33 2 02351 159 In 192 2 159 m 8025 0235 So in the year 2008 the population of Florida will reach 192 millions

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