Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises - Page 368: 44

Answer

$x=\ln5\approx1.609438$ and $x=\ln3\approx1.098612$

Work Step by Step

$e^{x}+15e^{-x}-8=0$ Rewrite the second term of this equation as $\dfrac{15}{e^{x}}$: $e^{x}+\dfrac{15}{e^{x}}-8=0$ Multiply the whole equation by $e^{x}$: $e^{x}\Big(e^{x}+\dfrac{15}{e^{x}}-8=0\Big)$ $e^{2x}+15-8e^{x}=0$ $e^{2x}-8e^{x}+15=0$ Factor this equation: $(e^{x}-5)(e^{x}-3)=0$ Set both factors equal to $0$ and solve each individual equation: $e^{x}-5=0$ Take the $-5$ to the right side: $e^{x}=5$ Apply $\ln$ to both sides: $\ln e^{x}=\ln5$ Take the exponent $x$ to multiply in front of the $\ln$: $x\ln e=\ln5$ Since $\ln e=1$, the solution is: $x=\ln5\approx1.609438$ Let's move on to the other equation: $e^{x}-3=0$ Take the $-3$ to the right side of the equation: $e^{x}=3$ Apply $\ln$ to both sides: $\ln e^{x}=\ln3$ Take the exponent $x$ down to multiply in front of its respective $\ln$: $x\ln e=\ln3$ Since $\ln e=1$, the solution is: $x=\ln3\approx1.098612$ The two solutions for this equations are: $x=\ln5\approx1.609438$ and $x=\ln3\approx1.098612$
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