HW13-solutions

# HW13-solutions - husain(aih243 – HW13 – Gilbert...

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Unformatted text preview: husain (aih243) – HW13 – Gilbert – (56215) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Rewrite 3 2 = 9 in equivalent logarithmic form. 1. log 10 9 = 3 2. log 9 3 = 2 3. log 3 9 = 2 correct 4. log 3 9 = − 2 5. log 3 1 9 = 2 Explanation: Taking logs to the base 3 of both sides we see that log 3 9 = log 3 3 2 = 2 log 3 3 . But log 3 3 = 1 , so log 3 9 = 2. 002 10.0 points Rewrite 6 log 3 x = − 4 in equivalent exponential form. 1. x 4 = 1 81 2. x 6 = 1 81 correct 3. x 4 = 13 4. x 6 = 81 5. x 6 = − 13 Explanation: By exponentiation to the base 3, 3 6 log 3 x = 1 81 . But 3 6 log 3 x = 3 log 3 x 6 = x 6 . Hence the exponential form of the given equa- tion is x 6 = 1 81 . 003 10.0 points Find the value of x when x = 3 4 parenleftBig 3 log 3 2 parenrightBig − 5 parenleftBig 8 − log 8 4 parenrightBig without using a calculator. 1. x = 5 16 2. x = − 5 16 3. x = − 1 4 4. x = 1 4 correct 5. x = 3 8 Explanation: Since 3 log 3 x = x, 8 − log 8 x = 1 x , we see that 3 log 3 2 = = 2 , husain (aih243) – HW13 – Gilbert – (56215) 2 while 8 − log 8 4 = = 1 4 . Consequently, x = 3 2 − 5 4 = 1 4 . 004 10.0 points Evaluate f ( x ) = 2(ln x ) 2 + 3 ln x at x = 3 e . 1. f (3 e ) = 2(ln 3) 2 − 7 ln 3 − 5 2. f (3 e ) = 3(ln 3) 2 +7 ln 3 + 5 3. f (3 e ) = 2(ln 3) 2 + ln 3 − 1 4. f (3 e ) = 3(ln 3) 2 − ln 3 + 1 5. f (3 e ) = 3(ln 3) 2 + ln 3 − 1 6. f (3 e ) = 2(ln 3) 2 − ln 3 + 1 7. f (3 e ) = 2(ln 3) 2 +7 ln 3 + 5 correct 8. f (3 e ) = 3(ln 3) 2 − 7 ln 3 − 5 Explanation: Since ln(3 e ) = ln 3 + ln e = ln3 + 1 , we see that f (3 e ) = 2(ln3 + 1) 2 + 3(ln 3 + 1) = 2 braceleftBig (ln3) 2 + 2 ln3 + 1 bracerightBig + 3 braceleftBig ln3 + 1 bracerightBig . Consequently, f (3 e ) = 2(ln3) 2 +7 ln3 + 5 . 005 10.0 points Which one of the following could be the graph of f ( x ) = − log 2 ( x − 1) when dashed lines indicates asymptotes? 1. 2 4 2 − 2 2. 2 4 2 − 2 3. 2 4 2 − 2 husain (aih243) – HW13 – Gilbert – (56215) 3 4. 2 4 2 − 2 5. 2 4 2 − 2 correct 6. 2 4 2 − 2 Explanation: Let’s first review some properties of log 2 ( x ) and − log 2 ( x ). Since log 2 (1) = 0, the graph of log 2 ( x ) has x-intercept at x = 1. On the other hand, log 2 ( x ) is defined only on (0 , ∞ ) and lim x → + log 2 ( x ) = −∞ , lim x →∞ log 2 ( x ) = ∞ , so x = 0 is a vertical asymptote. Thus the graph of log 2 ( x ) is 2 4 2 − 2 To get the graph of − log 2 ( x ) we simply ‘flip’ the one for log 2 ( x )over the x-axis, producing 2 4 2 − 2 To obtain the graph of y = − log 2 ( x − 1) from this last one all we have to do now is translate horizontally to the right by 1, pro- ducing 2 4 2 − 2 husain (aih243) – HW13 – Gilbert – (56215) 4 keywords: LogFunc, LogFuncExam, 006 10.0 points The graph 2 4 6 − 2 − 4 − 6 − 8 − 10 2 4 6 8 − 2 − 4 − 6 − 8 P passes through the point...
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HW13-solutions - husain(aih243 – HW13 – Gilbert...

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