Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 12 - Section 12.3 - Exponential Functions - Exercise Set: 36

Answer

$x=6$

Work Step by Step

$\Big(\dfrac{1}{9}\Big)^{x}=27^{2-x}$ Rewrite $27$ as $3^{3}$ and $9$ as $3^{2}$: $\Big(\dfrac{1}{3^{2}}\Big)^{x}=(3^{3})^{2-x}$ Take $3^{2}$ to the numerator by changing the sign of its exponent: $(3^{-2})^{x}=(3^{3})^{2-x}$ Multiply the exponents on both sides of the equation: $3^{-2x}=3^{6-3x}$ If $3^{-2x}=3^{6-3x}$, then $-2x=6-3x$: $-2x=6-3x$ Solve for $x$: $-2x+3x=6$ $x=6$
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