Stitz-Zeager_College_Algebra_e-book

Stitz-Zeager_College_Algebra_e-book

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Unformatted text preview: ter, we isolate the exponential and take logs to get 2x = ln(4), or x = ln(4) = ln(2). (Can 2 you explain the last equality using properties of logs?) As in the previous example, we need to be careful about choosing test values. Since ln(1) = 0, we choose ln 1 , ln 3 and ln(3). 2 2 1 Evaluating,6 we have r(ln 1 ) = ln 1 e2 ln( 2 ) − 4 ln 1 . Applying the Power Rule to the log 2 2 12 ln( 2 ) 2 1 1 in the exponent, we obtain ln e − 4 ln = ln 1 eln( 4 ) − 4 ln 2 . Using the inverse 2 1 1 1 15 1 1 properties of logs, this reduces to 4 ln 2 − 4 ln 2 = − 4 ln 2 . Since 1 < 1, ln 2 < 0 2 and we get r(ln 1 ) is (+). Continuing in this manner, we find r(x) < 0 on (0, ln(2)). The 2 calculator confirms that the graph of f (x) = xe2x is below the graph of g (x) = 4x on this intervals.7 1 2 1 2 (+) 0 (−) 0 (+) 0 ln(2) y = f (x) = xe2x and y = g (x) = 4x 6 7 A calculator can be used at this point. As usual, we proceed without apologies, with the analytical method. Note: ln(2) ≈ 0.693. 364 Exponential and Logarithmic Functions Example 6.3.3. Re...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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