How do you solve $\ln x = 9$?

Hint: The given problem deals with the use of logarithms. It focuses on the basic definition of the logarithm function. We can solve the problem easily by converting the given natural logarithmic form into exponential form. For this, we need to have knowledge of interconversion of logarithmic function to exponential function.

Complete step by step answer:
Given question requires us to solve the logarithmic equation $\ln x = 9$ to find the value of variable x.
Firstly, we must know the value of the base of natural logarithm. The base of the natural logarithm is $e$. So, we have, ${\log _e}x = 9$.
Now, ${\log _e}x = 9$ can be easily changed into exponential form by understanding the interconversion between the logarithmic and exponential function.
The exponential form ${a^n} = b$ where base $ = a$, exponent$ = n$, value$ = b$ is written as ${\log _a}b = n$ in logarithmic form, read as log of ‘b’ to base ‘a’ is equal to ‘n’.
Therefore, the logarithmic form, read as log of ‘b’ to base ‘a’ is equal to ‘n’ can be written as ${a^n} = b$ in exponential form.
So, ${\log _e}x = 9$ can be written in exponential form as ${e^9} = x$.
So, changing the sides of equation, we get,
$ \Rightarrow x = {e^9}$
So, the value of x in the logarithmic equation $\ln x = 9$ is ${e^9}$.

Note:
The conversion of logarithmic form to exponential form is an easy task keeping in mind the readily interconvertible of the forms for a given number. We can change it vice-versa as well.
We can simplify these types of equations using following properties:
1. $\log(a)^x=x \log (a)$
2. $ \log(mn)= \log(m) +\log(n)$
3. $ \log\left(\dfrac{m}{n}\right)= \log(m) -\log(n)$