How do you evaluate ${{e}^{\ln y}}$ ?

Hint: To evaluate ${{e}^{\ln y}}$ , let us consider $x={{e}^{\ln y}}$ . We will then have to take the natural logarithm on both sides which yields $\ln x=\ln \left( {{e}^{\ln y}} \right)$ . Now, using the identity$\ln {{x}^{a}}=a\ln x$ , we will get \[\ln x=\ln \left( y \right)\ln \left( e \right)\] . When we apply the identity, $\ln e=1$ , we can further simplify the expression to \[\ln x=\ln y\] . The last step is to apply the rule, if $\ln a=\ln b$ then $a=b$ , which will result in the required answer.

Complete step by step solution:
We have to evaluate ${{e}^{\ln y}}$ . Let us consider $x={{e}^{\ln y}}$ .
Now, let us take the natural logarithm on both sides. We will get
$\ln x=\ln \left( {{e}^{\ln y}} \right)$
We know that $\ln {{x}^{a}}=a\ln x$ . Hence, the above form can be written as
\[\ln x=\ln \left( y \right)\ln \left( e \right)\]
We know that the natural logarithm of e is 1, that is, $\ln e=1$ . Hence, the above form becomes
\[\ln x=\ln \left( y \right)\times 1=\ln y\]
Let us apply the rule, if $\ln a=\ln b$ then $a=b$ . Hence, we can write the above equation as
\[\begin{align}
  & \Rightarrow x=y \\
 & \Rightarrow {{e}^{\ln y}}=y \\
\end{align}\]
Hence the answer is y.

Note: Students have a chance of making mistakes when using identities. They may write $\ln {{x}^{a}}=x\ln a$ instead of $\ln {{x}^{a}}=a\ln x$ . Also, they may consider the value of $\ln e$ to be -1 or 0. Note that logarithm of any value is never a negative. We can also solve this problem using the identity ${{a}^{{{\log }_{a}}x}}=x$ . This is explained below.
 We know that natural logarithm is the logarithm to the base e. This can be expressed as $\ln ={{\log }_{e}}$ .
Therefore, we can write ${{e}^{\ln y}}$ as ${{e}^{{{\log }_{e}}y}}$ .
Now, when we use the identity ${{a}^{{{\log }_{a}}x}}=x$ here, we will get
 ${{e}^{{{\log }_{e}}y}}=y$ .