Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Section 3.4 - Exponential and Logarithmic Equations - Exercise Set: 33

Answer

$x \approx -1.14$

Work Step by Step

The base in the exponential equation is $e$, so take the natural logarithm on both sides to obtain $\ln{e^{1-5x}}=\ln{793}.$ Use the property $\ln{e^b}=b$ (where b=1-5x) on the left side to obtain $1-5x = \ln{793}.$ Solve for $x$: $1-5x = \ln{793} \\-5x=\ln{793} - 1 \\x=\dfrac{\ln{793}-1}{-5}.$ Use a calculator and round-off the answer to two decimal places to obtain $x \approx -1.14.$
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.