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: 28


$x \approx 1.69$

Work Step by Step

The base in the exponential equation is $19$, so take the natural logarithm on both sides to obtain $\ln{19^x}=\ln{143}.$ Use the power rule $\ln{b^x} = x \ln{b}$ to bring the exponent to the front: $x \ln{19} = \ln{143}.$ Divide both sides of the equation by $\ln{19}$ to obtain $x=\dfrac{\ln{143}}{\ln{19}}.$ Use a calculator and round-off the answer to two decimal places to obtain $x \approx 1.69.$
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