Does e^(-ln(x)) Simplify to 1/x?

[1s1]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

Thanks!

 

[STEMucator]

Homework Statement

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

Homework Equations

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 $ln$.