College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.2 - Exponential Functions - 4.2 Exercises - Page 410: 74

Answer

$x=-\frac{3}{5}$

Work Step by Step

We rewrite both sides, so that the base is $\frac{1}{e}$ on both sides. $e^{x-1}=(\frac{1}{e}^4)^{x+1}$ By the law of exponents: $(a^x)^y=a^{xy}$ and $a^x=\frac{1}{a}^{-x}$ Left side: $e^{x-1}=(\frac{1}{e}^{-1})^{x-1}=\frac{1}{e}^{-x+1}$ Right side: $((\frac{1}{e})^4)^{x+1}=(\frac{1}{e})^{4x+4}$ The rewritten equation is: $\frac{1}{e}^{-x+1}=\frac{1}{e}^{4x+4}$ Then, since the sides have the same base, we set the exponents equal, because the two sides are equal only if the powers are equal too: if $b^m=b^n$ and $b \ne0$ , $b \ne1$ then $m=n$ $-x+1=4x+4 \\1-4=4x+x$ $-3=5x \\\frac{-3}{5}=\frac{5x}{5}$ $-\frac{3}{5}=x$
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