College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.3 - Exponential Functions - 6.3 Assess Your Understanding - Page 436: 73



Work Step by Step

To solve the given equation, make the two sides have the same base. Note that $8 = 2^3$ and $16=2^4$, so the given equation is equivalent to: $(2^3)^{-x+14}=(2^4)^x$ Use the rule $(a^m)^n = a^{mn}$ to obtain: $2^{3(-x+14)} = 2^{4x} \\2^{-3x+42}=2^{4x}$ Use the rule $a^m=a^n \longrightarrow m=n$ to obtain: $-3x+42=4x$ Add $3x$ to both sides of the equation to obtain: $\begin{array}{ccc} &-3x+42+3x &= &4x+3x \\&42 &= &7x \end{array}$ Divide 7 on both sides of the equation to obtain: $\begin{array}{ccc} &\frac{42}{7} &= &\frac{7x}{7} \\&6 &= &x \end{array}$ Thus, the solution set is $\left\{6\right\}$.
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