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: 72


The solution set is $\left\{0, \frac{1}{2}\right\}$.

Work Step by Step

To solve the given equation, make the two sides have the same base. Note that $4 = 2^2$ so the given equation is equivalent to: $(2^2)^{x^2}=2^x$ Use the rule $(a^m)^n = a^{mn}$ to obtain: $2^{2x^2} = 2^{x}$ Use the rule $a^m=a^n \longrightarrow m=n$ to obtain: $2x^2 = x$ Subtract $x$ to both sides of the equation to obtain: $2x^2-x=0$ Factor out the GCF $x$ to obtain: $x(2x-1)=0$ Equate each factor to $0$, then solve each equation to obtain: $\begin{array}{ccc} &x=0 &\text{ or } &2x-1=0 \\&x=0 &\text{ or } &2x=1 \\&x=0 &\text{ or } &x = \frac{1}{2} \end{array}$ Thus, the solution set is $\left\{0, \frac{1}{2}\right\}$.
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