Inverse functions and logs -solutions

Inverse functions and logs -solutions - handa(nh5757 –...

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Unformatted text preview: handa (nh5757) – Inverse functions and logs – bormashenko – (54880) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the value of x when x = 4 3 parenleftBig 2 log 2 2 parenrightBig − 9 parenleftBig 3 − log 3 4 parenrightBig without using a calculator. 1. x = 1 2 2. x = − 5 12 3. x = 5 12 correct 4. x = 7 12 5. x = − 1 2 Explanation: Since 2 log 2 x = x, 3 − log 3 x = 1 x , we see that 2 log 2 2 = = 2 , while 3 − log 3 4 = = 1 4 . Consequently, x = 8 3 − 9 4 = 5 12 . 002 10.0 points Use properties of logs to simplify the ex- pression log a 32 + 4 5 log a 10 − 4 5 log a 5 + log a 1 2 9 5 as much as possible. 1. log a 16 correct 2. log a 2 − 6 3. log a 32 4. log a 64 5. 16 Explanation: By properties of logs the given expression can be rewritten as log a braceleftBig 2 5 · 10 4 5 5 4 5 · 2 · 2 4 5 bracerightBig = log a 16 . 003 10.0 points Rewrite 3 4 = 81 in equivalent logarithmic form. 1. log 81 3 = 4 2. log 3 81 = − 4 3. log 3 81 = 4 correct 4. log 10 81 = 3 5. log 3 1 81 = 4 Explanation: Taking logs to the base 3 of both sides we see that log 3 81 = log 3 3 4 = 4 log 3 3 . But log 3 3 = 1 , so log 3 81 = 4 . handa (nh5757) – Inverse functions and logs – bormashenko – (54880) 2 keywords: LogFunc, LogFuncExam, 004 10.0 points Find the value of x when x = 7 log 2 parenleftBig 1 4 parenrightBig + 6 log 4 64 . 1. x = 4 correct 2. x = − 5 3. x = 6 4. x = 5 5. x = − 4 6. x = − 6 Explanation: By properties of logs, log 2 parenleftBig 1 4 parenrightBig = log 2 parenleftBig 1 (2) 2 parenrightBig = − 2 , while log 4 64 = log 4 (4) 3 = 3 . Consequently, x = 18 − 14 = 4 . 005 10.0 points If f is a one-to-one function such that f (8) = 2, what is the value of f − 1 (2)? 1. f − 1 (2) = 10 2. not enough information given 3. f − 1 (2) = 8 correct 4. f − 1 (2) = 14 5. f − 1 (2) = 6 6. f − 1 (2) = 13 Explanation: Since f (8) = 2 and f is one-to-one, the inverse, f − 1 , of f exists and f − 1 ( f ( x )) = f ( f − 1 ( x )) = x . Consequently, f − 1 (2) = f − 1 ( f (8)) = 8 . keywords: inverse, one-to-one, 006 10.0 points Find the inverse function, f − 1 , of f ( x ) = 5 x √ 7 x 2 + 1 . 1. f − 1 ( x ) = √ 25 + 7 x 2 x 2. f − 1 ( x ) = x √ 25 + 7 x 2 3. f − 1 ( x ) = x √ 25 − 7 x 2 correct 4. f − 1 ( x ) = √ 25 − 7 x 2 7 x 5. f − 1 ( x ) = 7 √ 25 − x 2 6. f − 1 ( x ) = √ 25 − x 2 7 Explanation: The graph of f ( x ) = 5 x √ 7 x 2 + 1 , is shown as a continuous line in f f − 1 handa (nh5757) – Inverse functions and logs – bormashenko – (54880) 3 Since this graph passes the horizontal line test, f will have an inverse, f − 1 , whose graph is shown as a dotted line. Algebraically, this inverse function is defined by y = f − 1 ( x ) where x = 5 y radicalbig 7 y 2 + 1 ....
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This note was uploaded on 11/21/2011 for the course M 408N taught by Professor Gualdini during the Spring '10 term at University of Texas.

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Inverse functions and logs -solutions - handa(nh5757 –...

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