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: 78


The solution set is $\left\{-\frac{1}{3}, -1\right\}$.

Work Step by Step

To solve the given equation, make the two sides have the same base. Note that $9=3^2$ and $27=3^3$, so the given equation is equivalent to: $(3^2)^{2x} \cdot (3^3)^{x^2} = 3^{-1}$ Use the rule $(a^m)^n = a^{mn}$ to obtain: $3^{2(2x)} \cdot 3^{3(x^2)} = 3^{-1} \\3^{4x} \cdot 3^{3x^2}=3^{-1}$ Use the rule $a^m \cdot a^n = a^{m+n}$ to obtain: $3^{4x+3x^2} = 3^{-1} \\3^{3x^2+4x} = 3^{-1}$ Use the rule $a^m=a^n \longrightarrow m=n$ to obtain: $3x^2+4x=-1$ Add $1$ to both sides of the equation to obtain: $\begin{array}{ccc} &3x^2+4x +1 &= &-1+1 \\&3x^2+4x+1 &= &0 \end{array}$ Factor the trinomial to obtain: $(3x+1)(x+1)=0$ Equate each factor to zero, and then solve each equation to obtain: $\begin{array}{ccc} &3x+1=0 &\text{ or } &x+1-0 \\&3x=-1 &\text{ or } &x=-1 \\&x=-\frac{1}{3} &\text{ or } &x=-1 \end{array}$ Thus, the solution set is $\left\{-\frac{1}{3}, -1\right\}$.
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