Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.3 One-to-One Functions; Inverse Functions - 5.3 Assess Your Understanding - Page 282: 81

Answer

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

Work Step by Step

Simplify the right side of the equation to obtain: $e^{x^2} = \dfrac{e^{3x}}{e^2}$ Use the rule $\dfrac{a^m}{a^n} = a^{m-n}$ to obtain: $e^{x^2} =e^{3x-2}$ Use the rule $a^m=a^n \longrightarrow m=n$ to obtain: $x^2=3x-2$ Subtract $3x$ and add $2$ on both sides of the equation to obtain: $\begin{array}{ccc} &x^2-3x+2 &= &3x-2-3x+2 \\&x^2-3x+2 &= &0\end{array}$ Factor the trinomial to obtain: $(x-2)(x-1)=0$ Equate each factor to zero, and then solve each equation to obtain: $\begin{array}{ccc} &x-2=0 &\text{ or } &x-1=0 \\&x=2 &\text{ or } &x=1\end{array}$ Thus, the solution set is $\left\{1, 2\right\}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.