than 1. It means in particular that the function is decreasing and approaching the x


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1 6 Preclculus Review Grph the functions ) (/) ) log y = b y = Solution () The function y = is n eponentil function with bse smller thn It mens in prticulr tht the function is decresing nd pproching the  is when moves to the right Some vlues of y re listed below 9 4 y( ) = = 5, y( ) = = 5, y(0) =, y() = 07, y() = (b) This is typicl grph of logrithmic function The function is defined on (0, ), negtive on (0,), nd positive on (, ) Some vlues of the function re listed below:
2 log = 0, log =, log9 = log ( ) =, log (/ ) = log ( ) =, log (/ 9) = log ( ) = Solve equtions ) log (/) 5 ) + = b = 5 Solution () to solve this eqution we write log (/) 5 logrithms the left prt equls to / nd we hve = By the definition of 5 = = (b) Let us tke nturl logrithms of both prts of the eqution 5 =, whence + ln( ) ln(5 ) = By properties of logrithms we cn write the lst eqution s ( ) ln ( ) ln 5 = + After epnding the epressions in both prts we obtin (ln ) ln (ln 5) ln 5 hve = + nd therefore (ln ln 5) = ln 5 + ln Finlly we ln 5 + ln = 0 ln ln 5
3 Find the domin f b f ) ( ) = log( 4) ) ( ) = log ( 9) Solution () log( 4) is defined if 4> 0 It mens tht 4 > 4 nd > 6 Therefore the domin of f is 4, (b) For the squre root to be defined we needlog ( 9) 0 This inequlity will be stisfied if 9 or 0 0 It cn be written s ( + 0)( 0) 0 The lst inequlity holds if either 0 (both fctors in the left prt re nonpositive) or 0 (both fctors re nonnegtive) The domin is (, 0) ( 0, ) 4 Simplify 6 + y + y ) log ( ) log ( ) b) log ( / ) log Solution () Using the properties of logrithms nd the formul y ( y)( y y ) + = + + we cn write + y log ( + y ) log ( + y) = log = log ( y + y ) + y (b) First notice tht for epression to mke sense we must hve > 0 whence > 0 nd = Net, using properties of logrithms we write log ( / ) log = log log log log = = log log = log
4 5 Solve log ) 4 4 7, ) ( ) log = b + = log4 Solution () by the definition of logrithms we hve 4 ( ) log 4 log4 4 = 4 = nd our eqution becomes = 7 = whence Only the positive solution mkes sense (logrithm of negtive number is undefined) nd therefore = 7 (b) By the property of logrithms log = whence ( ) + = nd ( + ) = 0 Fctoring out we get [( + ) ] = 0 One solution is = 0, we obtin two more solutions from the qudrtic eqution + = 0 Solving by qudrtic formul we get ± ± = = the originl eqution 4 ( ) All three solutions mke sense nd stisfy 6 Solve + 4 = 7 Solution = = = whence 4 + =, + 4+ = 0, ( + )( + ) = 0, nd we hve two solutions, 7 Solve e = = + = 5e Solution becuse e = ( e ) we hve qudrtic eqution for e : ( e ) 5e + = 0 Solving for e by qudrtic formul we get e 5 ± = Both solutions re positive nd therefore mke sense (recll tht e > 0 ) Tking nturl logrithms of both prts we get two solutions of the originl eqution
5 5 5 + = ln 57 nd = ln 57 8 Solve ) log( ) = log b) log( ) log( ) = Solution () becuse log( ) log = our eqution cn be written s log log 0 = or log log = 0 If we tke u = log we get qudrtic eqution for u: u u = 0or ( ) 0 uu = It hs two solutions u 0 nd u In the first cse log = 0 whence log = 0 In the second cse = = log = whence log = 4 Respectively we hve two solutions for : 0 4 = 0 = nd = 0 = 0000 (b) By the properties of logrithms log = whence 9 = = = = = = 8 0 0, 0 0, 8 9, 75 9 The popultion of Dlls ws 680,000 in 960 In 969 it ws 85,000 Find the corresponding eponentil model of the popultion growth nd estimte the popultion in 00
6 Solution Let t = 0correspond to yer 960, then t = 9 corresponds to yer 969, nd t = 50 to yer 00 The eponentil model of popultion growth cn be written s Pt ( ) = P(0) e kt We know tht P (0) = nd tht P (9) = From it we cn find 6 e = =, 9k = ln tht 9 k, nd 6 ln k = Therefore our prediction 50k for yer 00 is P e e (50) = The hlflife of Polonium is min After 8 min how much of 40 g smple remins rdioctive? Solution The eqution of rdioctive decy is Qt ( ) = Q(0) e kt, where in our cse t is time in minutes nd Q (0) = 40 Becuse hlflife of Polonium is min we hve Q() Q(0) ln = whence k e =, k = ln = ln, nd k = 00 8k 8 ( 0) Finlly, Q(8) = Q(0) e 40 e 66g = 66mg
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