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: 41

Answer

$x \approx -2.80$

Work Step by Step

Take the natural logarithm on both sides to obtain $\ln{(5^{2x+3})}=\ln{(3^{x-1})}.$ Use the power rule $\ln{a^x}=x\ln{a}$ to bring down the exponents: $(2x+3) \cdot \ln{5} = (x-1) \cdot \ln{3}.$ Use the distributive property to obtain $2x\ln{5}+3\ln{5}=x\ln{3}-\ln{3}.$ Combine like terms. Put all terms with the variable $x$ on the left side and the rest of the terms on the right side. Note that when terms move from one side to the other, the operation changes to its opposite. $2x\ln{5} - x\ln{3} = -\ln{3}-3\ln{5}.$ Factor out $x$ on the left side to obtain $x(2\ln{5}-\ln{3})=-\ln{3}-3\ln{5}.$ Divide both sides by $2\ln{5}-\ln{3}$ to obtain $x = \dfrac{-\ln{3}-3\ln{5}}{2\ln{5}-\ln{3}}.$ Use a calculator to find the value of $x$. Round-off the answers to two decimal places. $x \approx -2.80.$
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