Stitz-Zeager_College_Algebra_e-book

# Buer solutions have a wide variety of applications

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Unformatted text preview: e they intersect only ln(117) at x = −4. To see what happened to the solution x = 1, we substitute it into our original equation to obtain log117 (−2) = log117 (−2). While these expressions look identical, neither is a real number,1 which means x = 1 is not in the domain of the original equation, and is not a solution. 2. Our ﬁrst objective in solving 2 − ln(x − 3) = 1 is to isolate the logarithm. We get ln(x − 3) = 1, which, as an exponential equation, is e1 = x − 3. We get our solution x = e + 3. On the calculator, we see the graph of f (x) = 2 − ln(x − 3) intersects the graph of g (x) = 1 at x = e + 3 ≈ 5.718. 1 They do, however, represent the same family of complex numbers. We stop ourselves at this point and refer the reader to a good course in Complex Variables. 6.4 Logarithmic Equations and Inequalities y = f (x) = log117 (1 − 3x) and y = g (x) = log117 x2 − 3 369 y = f (x) = 2 − ln(x − 3) and y = g (x) = 1 3. We can start solving log6 (x + 4) + log6 (3 − x) = 1 by using the Product Rule for logarithms to rewrite the equation as log6 [(x...
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