Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Section 8.3 - Matrix Operations and Their Applications - Exercise Set - Page 920: 85

Answer

The solution set is $\left\{ \frac{11}{2} \right\}$

Work Step by Step

Consider the given system of equations: ${{3}^{2x-8}}=27$ In the given equation, we will express both sides as a power of the same base. ${{3}^{2x-8}}={{3}^{3}}$ If the bases are the same and not equal to unity, then the powers are equal. This implies, $2x-8=3$ Solve for x as below: $\begin{align} & 2x-8=3 \\ & 2x-8+8=3+8 \\ & 2x=11 \end{align}$ Dividing both sides by 2, we get, $\begin{align} & 2x=11 \\ & \frac{2x}{2}=\frac{11}{2} \\ & x=\frac{11}{2} \end{align}$ Now, substituting the value $\frac{11}{2}$ for $x$ into the original equation, we will check the solution, $\begin{align} & {{3}^{2x-8}}=27 \\ & {{3}^{2\left( \frac{11}{2} \right)-8}}=27 \\ & {{3}^{3}}=27 \\ & 27=27 \end{align}$
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