Stitz-Zeager_College_Algebra_e-book

If we are to attack this data from a scientic

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Unformatted text preview: to inequalities involving logarithmic functions. Since logarithmic functions are continuous on their domains, we can use sign diagrams. Example 6.4.2. Solve the following inequalities. Check your answer graphically using a calculator. 2 Recall that an extraneous solution is an answer obtained analytically which does not satisfy the original equation. 6.4 Logarithmic Equations and Inequalities 1. 1 ≤1 ln(x) + 1 371 2. (log2 (x))2 < 2 log2 (x) + 3 3. x log(x + 1) ≥ x Solution. 1. We start solving 1 ln(x)+1 1 ln(x)+1 − 1 − reduces to ln(ln(x) x)+1 ≤ 1 by getting 0 on one side of the inequality: 1 Getting a common denominator yields ln(x)+1 ln(x) ln(x) or ln(x)+1 ≥ 0. We define r(x) = ln(x)+1 and − ln(x)+1 ln(x)+1 ≤ 0 which ≤ 0. ≤ 0, set about finding the domain and the zeros of r. Due to the appearance of the term ln(x), we require x > 0. In order to keep the denominator away from zero, we solve ln(x) + 1 = 0 so ln(x) = −1, so x = e−1 = 1 . Hence, e ln(x) the domain of r is 0, 1 ∪ 1 , ∞ . To ...
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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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