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: 75


The solution set is $\left\{-1, 7\right\}$.

Work Step by Step

To solve the given equation, make the two sides have the same base. Note that $27=3^3$, so the given equation is equivalent to: $3^{x^2-7}=(3^3)^{2x}$ Use the rule $(a^m)^n = a^{mn}$ to obtain: $3^{x^2-7} = 3^{3(2x)} \\3^{x^2-7}=3^{6x}$ Use the rule $a^m=a^n \longrightarrow m=n$ to obtain: $x^2-7=6x$ Subtract $6x$ to both sides of the equation to obtain: $\begin{array}{ccc} &x^2-7-6x &= &6x-6x \\&x^2-6x-7 &= &0 \end{array}$ Factor the trinomial to obtain: $(x-7)(x+1)=0$ Equate each factor to zero, and then solve each equation to obtain: $\begin{array}{ccc} &x-7=0 &\text{ or } &x+1-0 \\&x=7 &\text{ or } &x=-1 \end{array}$ Thus, the solution set is $\left\{-1, 7\right\}$.
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