Stitz-Zeager_College_Algebra_e-book

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Unformatted text preview: Power and Quotient Rules, we obtain 1 = log2 x2 x+1 . x2 Rewriting this in exponential form, we get x+1 = 2 or x2 − 2x − 2 = 0. Using the quadratic √ ln(x formula, we get x = 1 ± 3. Graphing f (x) = 1 + 2 ln(x+1) and g (x) = 2ln(2)) , we see the ln(4) √ √ x graphs intersect only at x = 1 + 3 ≈ 2.732. The solution√ = 1 − 3 < 0, which means if substituted into the original equation, the term 2 log2 1 − 3 is undeﬁned. y = f (x) = 1 + 2 log4 (x + 1) and y = g (x) = 2 log2 (x) If nothing else, Example 6.4.1 demonstrates the importance of checking for extraneous solutions2 when solving equations involving logarithms. Even though we checked our answers graphically, extraneous solutions are easy to spot - any supposed solution which causes a negative number inside a logarithm needs to be discarded. As with the equations in Example 6.3.1, much can be learned from checking all of the answers in Example 6.4.1 analytically. We leave this to the reader and turn our attention...
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