Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 6 - Logarithmic Functions - 6.6 Solving Logarithmic Equations - 6.6 Exercises: 51

Answer

$t \approx 6.325$

Work Step by Step

$\log (5t^{3}) - \log (2t) = 2$ $\log \frac{5t^{3}}{2t} = 2$ $\log \frac{5t^{2}}{2} = 2$ $10^{2} = \frac{5t^{2}}{2}$ $100 = \frac{5t^{2}}{2}$ $200 = 5t^{2}$ $t^{2} = 40$ $t = ±\sqrt {40}$ $t \approx 6.325$ Since we can't take the log of a negative number, the only possible solution is $t = \sqrt {40}$. $t \approx 6.325$. Check: $\log (5(\sqrt {40})^{3}) - \log (2(\sqrt {40})) \overset{?}{=} 2$ $\log (5(252.982...) - \log (12.6491...) \overset{?}{=} 2$ $\log (1264.911...) - \log (12.6491...) \overset{?}{=} 2$ $\log \frac{(1264.911...)}{(12.6491...)} \overset{?}{=} 2$ $\log 100 \overset{?}{=} 2$ $ 2 = 2$
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