Algebra 2 Common Core

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

Chapter 11 - Probability and Statistics - 11-3 Probability of Multiple Events - Practice and Problem-Solving Exercises - Page 693: 59

Answer

$x=e^2$

Work Step by Step

Using the properties of logarithms, the given equation, $ \ln x^2+1=5 ,$ is equivalent to \begin{align*}\require{cancel} \ln x^2+1-1&=5-1 \\ \ln x^2&=4 \\ 2\ln x&=4 &\left(\text{use } \ln a^x=x\ln a \right) \\\\ \dfrac{2\ln x}{2}&=\dfrac{4}{2} \\\\ \ln x&=2 \\ \log_e x&=2 &\left(\text{use } \ln x=\log_ex \right) \\ x&=e^2 &\left(\text{use } y=b^x\Rightarrow\log_by=x \right) .\end{align*} Hence, the solution is $ x=e^2 .$ CHECKING: Substituting $ x=e^2 $ in the given equation, $ \ln x^2+1=5 ,$ results to \begin{align*} \ln (e^2)^2+1&=5 \\ \ln e^4+1=&5 &\left(\text{use } (a^m)^n=a^{mn} \right) \\ 4\ln e+1=&5 &\left(\text{use } \ln a^x=x\ln a \right) \\ 4(1)+1=&5 &\left(\text{use } \ln e=1 \right) \\ 4+1=&5 \\ 5=&5 &\text{ (TRUE)} .\end{align*} Since the substitution above ended with a TRUE statement, then $ x=e^2 $ is the solution.
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