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: 37


$x \approx 1.09$

Work Step by Step

The base in the exponential equation is $7$, so take the natural logarithm on both sides to obtain $\ln{7^{x+2}}=\ln{410}.$ Use the power rule $\ln{a^x}=x\ln{a}$ to bring down the exponent: $(x+2)\ln{7} = \ln{410}.$ Divide by $\ln{7}$ to obtain $x+2 = \dfrac{\ln{410}}{\ln{7}}.$ Subtract $2$ from both sides to obtain $x = \dfrac{\ln{410}}{\ln{7}}-2.$ Use a calculator and round-off the answer to two decimal places to obtain $x \approx 1.09.$
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