Answer
True.
Work Step by Step
For $a > 0, a\neq 1$, and $x > 0,\ \qquad y=\log_{a}x$ means $a^{y}=x$.
So, $a^{\log_{a}x}=x$
Also, for $a=e,\ \log_{a}x$ is written as $\ln x.$
Therefore, $e^{\ln 2}=2.$
The statement is true.
Calculus with Applications (10th Edition)
by Lial, Margaret L.; Greenwell, Raymond N.; Ritchey, Nathan P.
Published by
Pearson
ISBN 10:
0321749006
ISBN 13:
978-0-32174-900-0
Chapter 2 - Nonlinear Functions - Chapter Review - Review Exercises - Page 113: 14
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