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Unformatted text preview: montelongo (jcm3827) – hW 14 – gualdani – (56455) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Rewrite 2 − 4 = 1 16 in equivalent logarithmic form. 1. log 2 1 16 = 4 2. log 10 1 16 = 2 3. log 2 1 16 = 4 correct 4. log 2 16 = 4 5. log 1 16 2 = 4 Explanation: Taking logs to the base 2 of both sides we see that log 2 1 16 = log 2 2 − 4 = 4 log 2 2 . But log 2 2 = 1 , so log 2 1 16 = 4. 002 10.0 points Rewrite 7 log 3 x = 3 in equivalent exponential form. 1. x 7 = 27 2. x 3 = 10 3. x 7 = 10 4. x 7 = 1 27 correct 5. x 3 = 1 27 Explanation: By exponentiation to the base 3, 3 7 log 3 x = 1 27 . But 3 7 log 3 x = 3 log 3 x 7 = x 7 . Hence the exponential form of the given equa tion is x 7 = 1 27 . 003 10.0 points Use properties of logs to simplify the ex pression log 4 ( x radicalbig x 2 32 ) + log 4 ( x + radicalbig x 2 32 ) . 1. 1 + log 8 4 2. log 4 8 3. 4 + log 4 8 4. log 8 4 5. 1 + log 4 8 correct Explanation: By properties of logs the given expression can be rewritten as log 4 braceleftBig ( x radicalbig x 2 32 ) ( x + radicalbig x 2 32 ) bracerightBig = log 4 braceleftBig x 2 ( radicalbig x 2 32 ) 2 bracerightBig . montelongo (jcm3827) – hW 14 – gualdani – (56455) 2 Thus the given expression reduces to log 4 32 = 1 + log 4 8 since log 4 32 = log 4 4 + log 4 8 . 004 10.0 points Simplify the expression f ( x ) = 4 7(log 4 e ) ln x as much as possible. 1. f ( x ) = 7 x 2. f ( x ) = x 7 correct 3. f ( x ) = x 28 4. f ( x ) = x 4 5. f ( x ) = e 29 Explanation: By the property of inverse functions, 4 log 4 y = y, e ln y = y . Consequently, f ( x ) = 4 7(log 4 e ) ln x = (4 log 4 e ) 7 ln x = e ln x 7 = x 7 . 005 10.0 points Which one of the following could be the graph of f ( x ) = log 3 parenleftBig 1 x 2 parenrightBig when a dashed line indicates an asymptote? 1. 2. 3. 4. montelongo (jcm3827) – hW 14 – gualdani – (56455) 3 5. 6. cor rect Explanation: Let’s first review some properties of ln x and ln( x ). Since ln x is defined only on (0 , ∞ ) and lim x → + ln x =∞ , lim x →∞ ln x = ∞ , the graph of ln x has a vertical asymptote at x = 0 and so is given by But then ln( x ) is defined only on (∞ , 0) and has the properties lim x → ln( x ) =∞ , lim x →−∞ ln( x ) = ∞ , so its graph has a vertical asymptote at x = 0 and is given by Now the given function is f ( x ) = log 3 parenleftBig 1 x 2 parenrightBig = log 3 ( x 2) . Its graph will have a vertical asymptote at x = 2, and so will be that of log 3 ( x ) translated 2 units to the left, then ‘flipped over’ the x axis. Consequently, f has graph keywords: LogFunc, LogFuncExam, 006 10.0 points Which of the following is the graph of the function y = 1 log 2 ( x + 8)?...
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 Spring '10
 Gualdani
 Calculus, Derivative, Limit of a function, Logarithm

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