Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 11 - Probability and Statistics - 11-4 Conditional Probability - Practice and Problem-Solving Exercises - Page 702: 45

Answer

$x\approx3.465$

Work Step by Step

Using the properties of equality, the given equation, $ 7-3^x=-38 ,$ is equivalent to \begin{align*} 7+38&=3^x \\ 45&=3^x .\end{align*} Taking the logarithm of both sides, the equation above is equivalent to \begin{align*} \log45&=\log3^x .\end{align*} Using the properties of logarithms, the equation above is equivalent to \begin{align*} \log45&=x\log3 &\left(\text{use }\log_ba^x=x\log_ba \right) \\\\ \dfrac{\log45}{\log3}&=\dfrac{x\log3}{\log3} \\\\ \dfrac{\log45}{\log3}&=x \\\\ x&\approx3.465 .\end{align*} Hence, the solution is $ x\approx3.465 .$
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