## College Algebra (11th Edition)

Published by Pearson

# Chapter 4 - Review Exercises - Page 469: 41

#### Answer

$x\approx-0.485$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $e^{2-x}=12 ,$ take the natural logarithm of both sides. Then use the laws of logarithms and the properties of equality to isolate the variable. Express the answer with $3$ decimal places. $\bf{\text{Solution Details:}}$ Taking the natural logarithm of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \ln e^{2-x}=\ln 12 .\end{array} 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} (2-x)\ln e=\ln 12 .\end{array} Since $\ln e =1,$ the equation above is equivalent to \begin{array}{l}\require{cancel} (2-x)(1)=\ln 12 \\\\ 2-x=\ln 12 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} -x=-2+\ln 12 \\\\ -1(-x)=-1(-2+\ln 12) \\\\ x=2-\ln 12 \\\\ x\approx-0.485 .\end{array}

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