College Algebra (11th Edition)

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

Chapter 4 - Test: 15

Answer

$x\approx2.811$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ e^{0.4x}=4^{x-2} ,$ take the natural logarithm of both sides. Use the properties of logarithms and of equality to isolate the variable. Approximate the answer with $3$ decimal places. $\bf{\text{Solution Details:}}$ Taking the natural logarithm of both sides results to \begin{array}{l}\require{cancel} \ln e^{0.4x}=\ln 4^{x-2} .\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} 0.4x\ln e=(x-2)(\ln 4) .\end{array} Since $\ln e=1,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 0.4x(1)=(x-2)(\ln 4) \\\\ 0.4x=(x-2)(\ln 4) .\end{array} Using the Distributive Property and the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 0.4x=x(\ln 4)-2(\ln 4) \\\\ 0.4x=x\ln 4-2\ln 4 \\\\ 0.4x-x\ln 4=-2\ln 4 \\\\ x(0.4-\ln 4)=-2\ln 4 \\\\ x=\dfrac{-2\ln 4}{0.4-\ln 4} \\\\ x\approx2.811 .\end{array}
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