How do you simplify $ {{e}^{-\ln x}} $ ?

Hint:
The mathematical operation "ln" stands for natural logarithm, i.e. log to the base e", where e is the Euler's constant. If $ {{e}^{x}} $ = b, then we write "ln b = x". Assume the given expression to be equal to some variable y and simplify using the properties of log.
It is worth recalling that $ \log {{a}^{x}} $ = x log a.

Complete Step by step Solution:
Let's say that $ {{e}^{-\ln x}} $ = y. Using the definition of the log function, we know that this is the same as:
⇒ ln y = − ln x
Using the property of logarithms $ \log {{a}^{x}} $ = x log a, we get:
⇒ ln y = $ \ln \left( {{x}^{-1}} \right) $
⇒ ln y = $ \ln \left( \dfrac{1}{x} \right) $
Since, for real numbers, logs of equal numbers are equal, we can say that:
⇒ y = $ \dfrac{1}{x} $
According to our assumption, $ {{e}^{-\ln x}} $ = y, therefore:

⇒ $ {{e}^{-\ln x}} $ = $ \dfrac{1}{x} $, which is the required simplification.

Note:
If $ {{a}^{x}} $ = b, then we write $ {{\log }_{a}}b $ = x. The base "a" can be any real number except 0 and 1. The Euler's constant 'e' is a useful number when we encounter powers. It's an irrational number and its value is e = $ \underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\dfrac{1}{x} \right)}^{x}} $ = 2.7182...
Some rules of logarithms:
log a + log b = log (ab)
log a − log b = $ \log \left( \dfrac{a}{b} \right) $
 $ \log {{a}^{x}} $ = x log a