Whereas the radioactive decay 10 the time it takes

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Unformatted text preview: ) − 4 3−4 e We determine the signs of r (ln(5)) and r (ln(7)) similarly.5 From the sign diagram, we find our answer to be (−∞, ln(4)) ∪ [ln(6), ∞). Using the calculator, we see the graph of ex f (x) = ex −4 is below the graph of g (x) = 3 on (−∞, ln(4)) ∪ (ln(6), ∞), and they intersect at x = ln(6) ≈ 1.792. 4 This is because the base of ln(x) is e > 1. If the base b were in the interval 0 < b < 1, then logb (x) would decreasing. 5 We could, of course, use the calculator, but what fun would that be? 6.3 Exponential Equations and Inequalities (−) 363 (+) 0 (−) ln(4) ln(6) y = f (x) = ex ex −4 y = g (x) = 3 3. As before, we start solving xe2x < 4x by getting 0 on one side of the inequality, xe2x − 4x < 0. We set r(x) = xe2x − 4x and since there are no denominators, even-indexed radicals, or logs, the domain of r is all real numbers. Setting r(x) = 0 produces xe2x − 4x = 0. With x both in and out of the exponent, this could cause some difficulty. However, before panic sets in, we factor out the x to obtain x e2x − 4 = 0 which gives x = 0 or e2x − 4 = 0. To solve the lat...
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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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