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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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