# E^-ln(x) = 1/x?

## Homework Statement

Is $e^{-ln(x)}$ equal to $\frac{1}{x}$ ?

## The Attempt at a Solution

$e^{-ln(x)} = e^{ln(x^{-1})} = e^{ln(\frac{1}{x})} = \frac{1}{x}$ ?

Thanks!

Related Calculus and Beyond Homework Help News on Phys.org
Zondrina
Homework Helper

## Homework Statement

Is $e^{-ln(x)}$ equal to $\frac{1}{x}$ ?

## The Attempt at a Solution

$e^{-ln(x)} = e^{ln(x^{-1})} = e^{ln(\frac{1}{x})} = \frac{1}{x}$ ?

Thanks!
Yes that is correct. It's a very common mistake that I see to not apply the log rules before eliminating $e$ and $ln$.