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This makes sense because after all the denition of

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Unformatted text preview: to get 6. What are we doing with this exponent? We are putting it on 117. By definition we get 6. In other words, the exponential function f (x) = 117x undoes the logarithmic function g (x) = log117 (x). Up until this point, restrictions on the domains of functions came from avoiding division by zero and keeping negative numbers from beneath even radicals. With the introduction of logs, we now have another restriction. Since the domain of f (x) = logb (x) is (0, ∞), the argument8 of the log must be strictly positive. Example 6.1.4. Find the domain of the following functions. Check your answers graphically using the calculator. 1. f (x) = 2 log(3 − x) − 1 2. g (x) = ln x x−1 Solution. 1. We set 3−x > 0 to obtain x < 3, or (−∞, 3). The graph from the calculator below verifies this. Note that we could have graphed f using transformations. Taking a cue from Theorem 1.7, we rewrite f (x) = 2 log10 (−x + 3) − 1 and find the main function involved is y = h(x) = log10 (x). 1 We select three points to track, 10 , −1 ,...
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