College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises: 45

Answer

$x=e$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \ln x+\ln x^2=3 ,$ use the properties of logarithms to simplify the given expression. Then convert to exponential form. $\bf{\text{Solution Details:}}$ Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} \ln x+2\ln x=3 .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (1+2)\ln x=3 \\\\ 3\ln x=3 \\\\ \ln x=\dfrac{3}{3} \\\\ \ln x=1 .\end{array} Since $\ln x=\log_e x,$ the equation above is equivalent to \begin{array}{l}\require{cancel} \log_e x=1 .\end{array} Since $y=b^x$ is equivalent to $\log_b y=x,$ the exponential form of the equation above is \begin{array}{l}\require{cancel} e^1=x \\\\ x=e .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.