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 - Page 861: 35



Work Step by Step

$\Big(\dfrac{1}{8}\Big)^{x}=16^{1-x}$ Rewrite $8$ as $2^{3}$ and $16$ as $2^{4}$: $\Big(\dfrac{1}{2^{3}}\Big)^{x}=(2^{4})^{1-x}$ Take $2^{3}$ to the numerator by changing the sign of its exponent: $(2^{-3})^{x}=(2^{4})^{1-x}$ Multiply the exponents on both sides: $2^{-3x}=2^{4-4x}$ If $2^{-3x}=2^{4-4x}$, then $-3x=4-4x$: $-3x=4-4x$ Solve for $x$: $-3x+4x=4$ $x=4$
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