## College Algebra (10th Edition)

Published by Pearson

# Chapter 6 - Section 6.3 - Exponential Functions - 6.3 Assess Your Understanding - Page 436: 77

#### Answer

The solution set is $\left\{-4, 2\right\}$.

#### Work Step by Step

To solve the given equation, make the two sides have the same base. Note that $4=2^2$ and $16=2^4$, so the given equation is equivalent to: $(2^2)^x \cdot 2^{x^2} = (2^4)^2$ Use the rule $(a^m)^n = a^{mn}$ to obtain: $2^{2x} \cdot 2^{x^2} = 2^{4(2)} \\2^{2x} \cdot 2^{x^2}=2^{8}$ Use the rule $a^m \cdot a^n = a^{m+n}$ to obtain: $2^{2x+x^2} = 2^8 \\2^{x^2+2x} = 2^8$ Use the rule $a^m=a^n \longrightarrow m=n$ to obtain: $x^2+2x=8$ Subtract $8$ to both sides of the equation to obtain: $\begin{array}{ccc} &x^2+2x -8 &= &8-8 \\&x^2+2x-8 &= &0 \end{array}$ Factor the trinomial to obtain: $(x+4)(x-2)=0$ Equate each factor to zero, and then solve each equation to obtain: $\begin{array}{ccc} &x+4=0 &\text{ or } &x-2-0 \\&x=-4 &\text{ or } &x=2 \end{array}$ Thus, the solution set is $\left\{-4, 2\right\}$.

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