College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Chapter 4 Review - Concept Check - Page 425: 66


no solution

Work Step by Step

RECALL: (1) $\ln{P} + \ln{Q}=\ln{(PQ)}$ (2) $\ln{P}=\ln{Q} \longrightarrow P=Q$ Use rule (1) above to obtain $\ln{[(x-2)(3)]}=\ln{(5x-7)}$ Use rule (2) above to obtain: $(x-2)(3)=5x-7 \\x(3) -2(3) = 5x-7 \\3x-6=5x-7$ Add $7$ and subtract$3x$ on both sides of the equation to obtain: $3x-6+7-3x=5x-7+7-3x \\(3x-3x) -6+7=(5x-3x) -7+7 \\1=2x$ Divide $2$ on both sides of the equation to obtain: $\frac{1}{2}=x$ However, ,if $x=\frac{1}{2}$, $\ln{(x-2)} = \ln{(\frac{1}{2}-2)} = \ln{(-\frac{3}{2})}$ which is undefined. $\ln{(5-7)} = \ln{(5\cdot\frac{1}{2} - 7)} = \ln{(\frac{5}{2} - \frac{14}{2})} = \ln{(-\frac{9}{2})}$ which is undefined. Thus, the given equation has no solution.
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