Stitz-Zeager_College_Algebra_e-book

At this point even though the function is still

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Unformatted text preview: (p) x = ln 4 5 (o) t = 2 ln 377 −0.8 (e) −∞, (f) 1 ln 18 ,∞ −0.1 = −∞, − 5 ln 4 2 377 = [10 ln(18), ∞) (d) (−∞, 1] (e) (−∞, 2.7095) (f) x ≈ 0.01866, x ≈ 1.7115 4. x > 1 (ln(6) + 1) 3 5. f −1 = ln x + √ x2 + 1 . Both f and f −1 have domain (−∞, ∞) and range (−∞, ∞). 368 6.4 Exponential and Logarithmic Functions Logarithmic Equations and Inequalities In Section 6.3 we solved equations and inequalities involving exponential functions using one of two basic strategies. We now turn our attention to equations and inequalities involving logarithmic functions, and not surprisingly, there are two basic strategies to choose from. For example, suppose we wish to solve log2 (x) = log2 (5). Theorem 6.4 tells us that the only solution to this equation is x = 5. Now suppose we wish to solve log2 (x) = 3. If we want to use Theorem 6.4, we need to rewrite 3 as a logarithm base 2. We can use Theorem 6.3 to do just that: 3 = log2 23 = log2 (8). Our equation then becomes log2 (x) = log2 (8) so that x = 8. However, we could have arrived at the same answer, in fewer steps, by using Theorem 6.3 to rewrite the equation log2 (x) = 3 as 23 = x, or x =...
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