Stitz-Zeager_College_Algebra_e-book

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Unformatted text preview: + 4)(3 − x)] = 1. Rewriting this as an exponential equation, we get 61 = (x + 4)(3 − x). This reduces to x2 + x − 6 = 0, which gives x = −3 and x = 2. Graphing y = f (x) = ln(x+4) + ln(3−x) and y = g (x) = 1, we see they intersect twice, at ln(6) ln(6) x = −3 and x = 2. y = f (x) = log6 (x + 4) + log6 (3 − x) and y = g (x) = 1 4. Taking a cue from the previous problem, we begin solving log7 (1 − 2x) = 1 − log7 (3 − x) by first collecting the logarithms on the same side, log7 (1 − 2x) + log7 (3 − x) = 1, and then using the Product Rule to get log7 [(1 − 2x)(3 − x)] = 1. Rewriting this as an exponential equation gives 71 = (1 − 2x)(3 − x) which gives the quadratic equation 2x2 − 7x − 4 = 0. Solving, we find 1 x = − 2 and x = 4. Graphing, we find y = f (x) = ln(1−2x) and y = g (x) = 1 − ln(3−x) intersect ln(7) ln(7) 1 only at x = − 2 . Checking x = 4 in the original equation produces log7 (−7) = 1 − log7 (−1), which is a clear domain violation. 5. Starting with log2 (x + 3) = log2 (6 ...
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