Q:

How $\ln(x) = \frac{5}{6}$ is simplified to $x = e^{\frac{5}{6}}$?

Hello I am trying to understand how my professor simplified this:
$$\ln(x) = \frac{5}{6}$$
into this: 

$$x = e^{\frac{5}{6}}$$

A:

Since my comment was well-received:
I always think of $\ln(x)$ as the answer to the question, "What power do I have to raise $e$ to, to get $x$?" So if $\ln(x) = 5/6$, then "$5/6$ is the power you need to raise $e$ to, to get $x$" hence the translation.

In my experience, people who successfully deal with logarithms are successful at translating back and forth between logarithmic and exponential equations; I hesitate to write the general equation
$$\log_b(x) = y \text{ if and only if } b^y = x,$$
but it really is just the equation-version of the idea that $\log_b(x)$ is the answer to the question, "What power do I have to raise $b$ to, to get $x$ (or more conversationally, the thing inside the logarithm)?"
This all comes from the idea that $\log_b(x)$ is defined to be the inverse of the base-$b$ exponential function, $y = b^x$, if you remember that inverse functions simply switch inputs and outputs.
In the case of the exponential-logarithmic inverse, it's nice to think in terms of 
$$\text{base}^\text{exponent} = \text{result} \longleftrightarrow \log_{\text{base}}(\text{result}) = \text{exponent};$$
logarithmic functions take the output from raising something to a power, and give you back the power. In a sense, I'm suggesting you think in terms of units/labels for the pieces of the equation -- that way you'll never be caught plugging an exponent into a logarithm, because the 'units' don't match!

A:

The natural logarithm is the log in base $e$ where $e$ is the Euler number

$$\ln(x) := \log_e(x)$$

You have the definition of logarithm and you can just apply to it.

$$\log_\color{blue}e(\color{red}x) = y \Leftrightarrow \color{blue}e^y = \color{red}x$$

