## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 12 - Exponential Functions and Logarithmic Functions - 12.6 Solving Exponential Equations and Logarithmic Equations - 12.6 Exercise Set: 28

#### Answer

$x=0$

#### Work Step by Step

Using the properties of equality, the given equation, $4+5e^{-x}=9 ,$ is equivalent to \begin{array}{l}\require{cancel} 5e^{-x}=9-4 \\\\ 5e^{-x}=5 \\\\ e^{-x}=\dfrac{5}{5} \\\\ e^{-x}=1 .\end{array} Taking the natural logarithm of both sides and using the properties of logarithms, the value of the variable that satisfies the equation, $e^{-x}=1 ,$ is \begin{array}{l}\require{cancel} \ln e^{-x}=\ln 1 \\\\ -x(\ln e)=\ln 1 \\\\ -x(1)=\ln 1 \\\\ -x=\ln 1 \\\\ x=-\ln 1 \\\\ x=0 .\end{array}

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