## College Algebra (11th Edition)

Published by Pearson

# Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises - Page 447: 38

#### Answer

$x=\dfrac{e^5}{2}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $\ln(2x)=5 ,$ convert it to exponential form. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Since $\ln x=\log_e x,$ the equation above is equivalent to \begin{array}{l}\require{cancel} \log_e(2x)=5 .\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^5=2x .\end{array} Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} 2x=e^5 \\\\ x=\dfrac{e^5}{2} .\end{array}

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