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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
formula, we get x = 1 ± 3. Graphing f (x) = 1 + 2 ln(x+1) and g (x) = 2ln(2)) , we see the
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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