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: 69


$x=\sqrt 2,-\sqrt 2,0$

Work Step by Step

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