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Name: Grade: out of 20 Intro to Calculus - Summer 2009 Instructor - Mike Limarzi Daily Quiz 10 - Thursday, July 16, 2009 1. (5 points) Rewrite as a sum or difference of multiples of logarithms. √ 3 xy . ln t4/3 √ 3 xy √ = ln 3 xy − ln(t4/3 ) ln t4/3 4 4 = ln (xy)1/3 − ln t = ln(x)1/3 + ln(y)1/3 − ln t 3 3 1 1 4 = ln x + ln y − ln t . 3 3 3 2. (5 points) Rewrite as a single expression. 3 log4 (x2 ) − 4 log4 (x−3 ) + 2 log4 (x). 3 log4 (x2 ) − 4 log4 (x−3 ) + 2 log4 (x) = log4 (x6 ) + log4 (x12 ) + log4 (x2 ). = log4 (x20 ) . 3. (5 points) Solve the following equation. log3 (x) = log3 2 − log3 (x − 2). 2 log3 (x) = log3 . Now raise 3 to the both sides: x−2 2 x = x−2 . We cross-multiply to get: 2 x − 2x = 2. Now, we bring the 2 over and use the quadratic formula: x2 − 2x − 2 = 0. √ √ √ 2± 4+8 2±2 3 x= = = 1 ± 3. 2 2 √ But, we need x > 0 and x − 2 > 0. Therefore, 1 − 3 is not a valid solution. This means we √ have only one answer: 1 + 3 . 4. (5 points) Solve the following equation. 2x−1 3x+2 1 1 = . 2 4 Take log2 of both sides: 2x−1 3x+2 1 1 log2 = log2 . Use the ”slap-it-over” rule: 2 4 1 1 (2x − 1) log2 = (3x + 2) log2 . Rewrite 12 = 2−1 and 2 4 1 4 = 2−2 . (2x − 1) log2 (2−1 ) = (3x + 2) log2 (2−2 ). By properties of logs, we get: (2x − 1)(−1) = (3x + 2)(−2). Simplifying: −2x + 1 = −6x − 4. Solving: 4x = −5. So, our solution is x = − Extra Credit: Name a type of cheese. 5 . 4