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: 10

Answer

$x=\pm2$

Work Step by Step

$10^{2x^{2}-3}=10^{9-x^{2}}$ Simply use the one-to-one property to solve this equation. Make the exponents on both sides of the equation equal: $2x^{2}-3=9-x^{2}$ Take the $-x^{2}$ to the left side of the equation and take $-3$ to the right side: $2x^{2}+x^{2}=9+3$ Simplify both sides: $3x^{2}=12$ Take the $3$ to divide the right side: $x^{2}=\dfrac{12}{3}$ $x^{2}=4$ Take the square root of both sides of the equation: $\sqrt{x^{2}}=\sqrt{4}$ $x=\pm2$
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