Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.1 Exponential And Logarithmic Functions - Exercises Set 6.1 - Page 419: 19



Work Step by Step

Putting a base $e$ under each side of the equation gives $e^{ln(1/x)}=e^{-2}$, which equals $\frac{1}{x}=e^{-2}$. Rewriting the equation as $$\frac{1}{x}=\frac{1}{e^{2}}$$ allows cross multiplication to be performed. This yields $x=e^2$.
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