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: 42

Answer

$x=\dfrac{1}{2}$

Work Step by Step

$3^{4x}-3^{2x}-6=0$ Rewrite the first term of this equation as $(3^{2x})^{2}$: $(3^{2x})^{2}-3^{2x}-6=0$ Factor this equation: $(3^{2x}-3)(3^{2x}+2)=0$ Set both factors equal to $0$ and solve each individual equation: $3^{2x}+2=0$ Take the $2$ to the right side: $3^{2x}=-2$ Since no values of $x$ make this first equation true, it has no solution. Let's move on to the other one: $3^{2x}-3=0$ Take the $3$ to the right side: $3^{2x}=3$ Simply use the one-to-one property to solve this equation. $3^{2x}=3$ $2x=1$ Solve for $x$: $x=\dfrac{1}{2}$
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