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: 43

Answer

$x=1$

Work Step by Step

$2^{x}-10(2^{-x})+3=0$ Rewrite the second term of this equation as $-\dfrac{10}{2^{x}}$: $2^{x}-\dfrac{10}{2^{x}}+3=0$ Multiply the whole equation by $2^{x}$: $2^{x}\Big(2^{x}-\dfrac{10}{2^{x}}+3=0\Big)$ $2^{2x}-10+3(2^{x})=0$ $2^{2x}+3(2^{x})-10=0$ Factor this equation: $(2^{x}+5)(2^{x}-2)=0$ Set both factors equal to $0$ and solve each individual equation: $2^{x}+5=0$ Take the $5$ to the right side: $2^{x}=-5$ Since no value of $x$ makes this first equation true, it has no solution. Let's move on to the other one: $2^{x}-2=0$ Take the $-2$ to the right side: $2^{x}=2$ Simply use the one-to-one property to solve this equation: $2^{x}=2$ $x=1$
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