How do you solve $\log x+ \log 20=2$ ?

Hint: We are given an equation involving two logarithmic functions and a constant term. Thus, we must have basic knowledge of the logarithmic functions in order to solve this equation. Firstly, we will apply the basic properties of logarithm on the logarithmic terms on the left-hand side of the equation to convert them into a single term. Further, we shall take the antilogarithm on both sides to finally find the solution for $x$.

Complete step-by-step solution:
We are given the equation $\log x+\log 20=2$.
We know by the well-defined properties of logarithm that when the sum of two logarithmic functions is given, then it can also be written as the log of product of those two functions and if the logarithm of product of two log functions is given, then it can also be written as the sum of logarithm of those two functions.
It is expressed as $\log a+\log b=\log ab$ and $\log ab=\log a+\log b$ respectively.
Here, we have $a=x$ and $b=20$. Thus, we have
$\Rightarrow \log 20x=2$
Taking antilog on both sides, we get
$\Rightarrow 20x={{e}^{2}}$
Now, we shall divide both sides by 20,
$\Rightarrow x=\dfrac{{{e}^{2}}}{20}$
Therefore, the solution of $\log x+ \log 20=2$ is $x=\dfrac{{{e}^{2}}}{20}$.

Note: Whenever a log is written in any mathematical equation, it represents the logarithm with base of exponent, e. If the base of the logarithm is something else such as $10$ or some other constant, then it is specifically written in the equation along with the log. However, if the base is not specified then we take the base as e, by default.