Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.3 Exponential Functions - 4.3 Assess Your Understanding - Page 307: 73

Answer

$x=7,-1$

Work Step by Step

Re-write the given equation as: $(3^3)^{2x}=3^{x^2-7}$ or, $3^{6x}=3^{x^2-7}$ Use the rule power rule: $a^p=a^q$. We can see that the base $a=2$ is the same on both sides of the equation. So, the exponents will also be equal. This implies that $p=q$ Therefore, $x^2-7=6x \\ x^2-6x-7=0 \\ (x-7)(x+1)=0$ By the zero-product property, we have: $x=7,-1$
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