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: 71


The solution set is $\left\{-\sqrt2, 0, \sqrt2\right\}$.

Work Step by Step

To solve the given equation, make the two sides have the same base. Note that $9 = 3^2$ so the given equation is equivalent to: $3^{x^3} = (3^2)^x$ Use the rule $(a^m)^n = a^{mn}$ to obtain: $3^{x^3} = 3^{2x}$ Use the rule $a^m=a^n \longrightarrow m=n$ to obtain: $x^3 = 2x$ Subtract $2x$ to both sides of the equation to obtain: $x^3-2x=0$ Factor out the GCF $x$ to obtain: $x(x^2-2)=0$ Equate each factor to $0$, then solve each equation to obtain: $\begin{array}{ccc} &x=0 &\text{ or } &x^2-2=0 \\&x=0 &\text{ or } &x^2=2\end{array}$ Take the square root of both sides of the second equation obtain: $\begin{array}{ccc} &x=0 &\text{ or } &x=\pm \sqrt{2}\end{array}$ Thus, the solution set is $\left\{-\sqrt2, 0, \sqrt2\right\}$.
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