Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - Review Exercises - Page 393: 36

Answer

$x = 3.303$

Work Step by Step

$\ln(x) + \ln(x-3) = 0$ $ln(x(x-3)) = 0$ $ln(x^{2}- 3x) = 0$ $x^{2} - 3x = e^{0}$ $x^{2} - 3x = 1$ $x^{2} - 3x - 1 = 0$ Use the quadratic formula to find $x$ $= \frac{-(-3) ±\sqrt {(-3)^{2}-4(1)(-1)}}{2(1)}$ $= \frac{3 ±\sqrt {9+4}}{2}$ $= \frac{3 ±\sqrt {13}}{2}$ $x = 3.303, -0.303$ Since $x \gt 0$, then the only answer is $x = 3.303$
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