How do you evaluate ${2^{\ln \pi }}$?

Hint: We will first assume the given expression to be equal to x and then take logarithmic function on both the sides, then we will find the required value using modifications.

Complete step-by-step solution:
We are given that we are required to evaluate the given expression which is given by ${2^{\ln \pi }}$.
Let us assume that the given expression is equal to x.
Therefore, we have $x = {2^{\ln \pi }}$
Let us take the natural logarithmic function on both the sides of the above mentioned expression, then we will obtain the following expression with us:-
$ \Rightarrow \ln x = \ln {2^{\ln \pi }}$ …………(1)
Now, we also know that we have a formula given by the following expression:-
$ \Rightarrow \ln {a^x} = x\ln a$
Replacing a by 2 and b by $\ln \pi $ in the above expression, we will then obtain the following expression with us:-
$ \Rightarrow \ln {2^{\ln \pi }} = \ln \pi .\ln 2$
Putting this in equation number 1, we will then obtain the following expression:-
$ \Rightarrow \ln x = \ln \pi .\ln 2$ ………………(2)
Now, we also know that we have a formula given by the following expression:-
If we have the expression given by $\ln a = b$ with us, then we have: $a = {e^x}$
Replacing a by x and b by $\ln \pi .\ln 2$ in the above mentioned formula, we will then obtain the following expression with us:-
$ \Rightarrow x = {e^{\ln \pi .\ln 2}}$
Putting this in equation number 2, we will then obtain the following expression:-
$ \Rightarrow x = {e^{\ln \pi .\ln 2}}$

Thus, $x = {e^{\ln \pi .\ln 2}}$ is the required answer.

Note: The students must note that they must commit to memory the following formulas:-
If we have the expression given by $\ln a = b$ with us, then we have: $a = {e^x}$
$\ln {a^x} = x\ln a$
The students might make the mistake of combining the logarithmic functions of breaking it into or something, but we do not have any such formula which contains the product of two logarithmic functions. Therefore, we left the answer in such a state.
The students might also make the mistake, where they may interpret that $x = \pi \ln 2$ from $x = {e^{\ln \pi .\ln 2}}$ but this would have been true if the logarithmic in the power would have been on the whole function altogether.