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 412: 110



Work Step by Step

First, we have to find the function, by calculating the value of the base, $a\\$. If the function contains this point (-2,36), then $f(-2)=36\\$. This means: $a^{-2}=36\\$. By the law of exponents: $a^{-x}=(\frac{1}{a})^x$ Thus, the equation above is equivalent to: $(\frac{1}{a})^{2}=36\\$ We take the square root of both sides but take only the positive root since $a \gt0$: $\frac{1}{a}=\sqrt{36} \\\frac{1}{a}=6\\$ $a \cdot \frac{1}{a} = 6 \cdot a \\$ $1=6a \\$ $\frac{1}{6}=a$ Therefore, with $a=\frac{1}{6}$, the function is: $f(x)=(\frac{1}{6})^x$.
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