Stitz-Zeager_College_Algebra_e-book

5 and that the half life10 of iodine 131 is

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Unformatted text preview: axis. 362 Exponential and Logarithmic Functions (+) 0 (−) 0 (+) −1 4 y = r(x) = 2x 2 −3x − 16 x e 2. The first step we need to take to solve ex −4 ≤ 3 is to get 0 on one side of the inequality. To that end, we subtract 3 from both sides and get a common denominator ex ex − 4 ≤3 ex −3 ≤ 0 −4 3 (ex − 4) ex − ≤ 0 Common denomintors. ex − 4 ex − 4 12 − 2ex ≤0 ex − 4 ex x −2 We set r(x) = 12x −e and we note that r is undefined when its denominator ex − 4 = 0, or e4 when ex = 4. Solving this gives x = ln(4), so the domain of r is (−∞, ln(4)) ∪ (ln(4), ∞). To find the zeros of r, we solve r(x) = 0 and obtain 12 − 2ex = 0. Solving for ex , we find ex = 6, or x = ln(6). When we build our sign diagram, finding test values may be a little tricky since we need to check values around ln(4) and ln(6). Recall that the function ln(x) is increasing4 which means ln(3) < ln(4) < ln(5) < ln(6) < ln(7). While the prospect of determining the sign of r (ln(3)) may be very unsettling, remember that eln(3) = 3, so r (ln(3)) = 12 − 2eln(3) 12 − 2(3) = −6 = ln(3...
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