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 - Page 488: 39

Answer

$x \approx 11.48$

Work Step by Step

The base in the exponential equation is $7$, so take the natural logarithm on both sides to obtain $\ln{(7^{0.3x})}=\ln{813}.$ Use the power rule $\ln{a^x}=x\ln{a}$ to bring down the exponent: $(0.3x)\ln{7} = \ln{813}.$ Divide both sides by $\ln{7}$ to obtain $0.3x = \dfrac{\ln{813}}{\ln{7}}.$ Divide both sides by $0.3$ to obtain $x = \dfrac{\frac{\ln{813}}{\ln{7}}}{0.3}.$ Use a calculator and round-off the answer to two decimal places to obtain $x \approx 11.48.$
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