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: 53

Answer

$x \approx 6.347$

Work Step by Step

$\log_5 x + \log_5 (2x+7) - 3 =0$ $\log_5 x + \log_5 (2x+7) =3$ $\log_5 x(2x+7) = 3$ $\log_5 (2x^{2} + 7x) = 3$ $125 = 2x^{2} + 7x$ $2x^{2} + 7x - 125 = 0$ $x = \frac{-(7)±\sqrt {7^{2}-4(2)(-125)}}{2(2)}$ $x = \frac{-7±\sqrt {49-4(2)(-125)}}{4}$ $x = \frac{-7±\sqrt {49+1000}}{4}$ $x = \frac{-7±\sqrt {1049}}{4}$ Since we can't take the log of a negative number, the only possible solution is $x = \frac{-7+\sqrt {1049}}{4}$ $x \approx 6.347$. Check: $\log_5 (\frac{-7+\sqrt {1049}}{4}) + \log_5 (2(\frac{-7+\sqrt {1049}}{4})+7) - 3 \overset{?}{=}0$ $\log_5 (\frac{-7+\sqrt {1049}}{4}) + \log_5 (19.6941...) - 3 \overset{?}{=}0$ $\log_5 ((\frac{-7+\sqrt {1049}}{4}) \times (19.6941...)) - 3 \overset{?}{=}0$ $\log_5 (125) - 3 \overset{?}{=}0$ $\log_5 (5^{3}) - 3 \overset{?}{=}0$ $3\log_5 (5) - 3 \overset{?}{=}0$ $3(1) - 3 \overset{?}{=}0$ $3 - 3 \overset{?}{=}0$ $0 = 0$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.