Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Review Exercises - Page 638: 48

Answer

$\left\{2.042\right\}$

Work Step by Step

Taking the logarithm of both sides, the given equation, $ 3^x=9.42 $ is equivalent to \begin{align*}\require{cancel} \log3^x&=\log9.42 .\end{align*} Using the properties of logarithms, the equation above is equivalent to \begin{align*}\require{cancel} x\log3&=\log9.42 &(\text{use }\log_b x^y=y\log_b x) .\end{align*} Using the properties of equality, the equation above is equivalent to \begin{align*}\require{cancel} \dfrac{x\cancel{\log3}}{\cancel{\log3}}&=\dfrac{\log9.42}{\log3} \\\\ x&=\dfrac{\log9.42}{\log3} .\end{align*} Using a calculator, the approximate values of each logarithmic expression above are \begin{align*} \log9.42&\approx0.97405 \\ \log3&\approx0.47712 .\end{align*} Substituting the approximate values in $ x=\dfrac{\log9.42}{\log3} $, then \begin{align*} x&\approx\dfrac{0.97405}{0.47712} \\\\ x&\approx2.042 .\end{align*} Hence, the solution set to the equation $ 3^x=9.42 $ is $ \left\{2.042\right\} $.
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