Math111i_GroupQuiz4_Solutions

Math111i_GroupQuiz4_Solutions - 3 5 x + 750 = 1000 5 x =...

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Math 111I Sec 001 Group Quiz 4 Name Directions: Read each question carefully and answer in the space pro- vided. The use of calculators is allowed, but not necessary. There are 10 points possible. To receive partial credit on any problem work must be shown. 1. Simplify the expression. Leave only positive exponents, and evaluate where necessary u - 7 / 3 · m 3 ( umu 3 ) 1 · m - 2 Solutions. The simplification follows as u - 7 / 3 · m 3 ( umu 3 ) 1 · m - 2 = u - 7 / 3 u 4 · m 3 m 1+( - 2) = u - 7 / 3 u 4 · m 3 m - 1 = u - 7 / 3 - 4 · m 3 - ( - 1) = u - 19 / 3 · m 4 = m 4 u 19 / 3 2. Solve for x ; 10 x = 526 , 325. Solutions. We begin by taking the Logarithm of both sides of the equation, and the answer follows nicely. 10 x = 526325 log(10 x ) = log(526325) x = log(526325) x 5 . 7 3. Solve for x : log(5 x + 750) = 3
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Math 111I Sec 001 Group Quiz 4 Solutions. We begin by taking both sides of the equation and raising 10 to that power. So we have log(5 x + 750) = 3 10 log(5 x +750) = 10
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Unformatted text preview: 3 5 x + 750 = 1000 5 x = 250 x = 50 . 4. Find C and a such that the function f ( x ) = Ca x satises the given conditions. f (0) = 7 and for each unit increase in x , the output is multiplied by 3.4. Solutions. We build this problem up from x = 0. So we begin by noting that f (0) = 7, so we see that f (0) = Ca = C = 7 . So we see that C = 7. Now we have that f (1) = f (0) 3 . 4, which follows as f (1) = f (0) 3 . 4 = 7 3 . 4 Now we compute f (2) and we nd that f (2) = f (1) 3 . 4 = 7 3 . 4 3 . 4 = 7(3 . 4) 2 Before we jump to a conclusion here, we should try one more case. And we see that f (3) = f (2) 3 . 4 = 7(3 . 4) 2 3 . 4 = 7(3 . 4) 3 Thus we see that f ( x ) = 7(3 . 4) x ....
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This note was uploaded on 02/16/2011 for the course MATH 111 taught by Professor Hitchcock during the Fall '08 term at South Carolina.

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Math111i_GroupQuiz4_Solutions - 3 5 x + 750 = 1000 5 x =...

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