If we are given the logarithmic equation ${\log _{\sqrt 3 }}5 = a$ and ${\log _{\sqrt 3 }}2 = b$ then ${\log _{\sqrt 3 }}300 = $
$A)2(a + b)$
$B)2(a + b + 1)$
$C)2(a + b + 2)$
$D)a + b + 4$
$E)a + b + 1$

Hint: First, we will understand what the logarithmic operator represents in mathematics. A logarithm function or log operator is used when we have to deal with the powers and base of a number, to understand it better which is $\log {x^m} = m\log x$
It’s basically a problem based on the logarithmic properties. We will mainly use two properties of the logarithm to solve this problem.

Formula used:
> Using the logarithm law, $\log (ab) = \log a + \log b$
> ${\log _{{a^n}}}b = \dfrac{1}{n}{\log _a}b$ where n is any number that we can choose as the base of the log

Complete step-by-step solution:
Given that we have two logarithm values with the same base \[\sqrt 3 \] and even the required function needs the same base and so it is easy to solve with the log properties.
Since we have ${\log _{\sqrt 3 }}5 = a$ and ${\log _{\sqrt 3 }}2 = b$. By using these two functions we need to find the value of ${\log _{\sqrt 3 }}300$
Let us convert the log function, ${\log _{\sqrt 3 }}300 = {\log _{\sqrt 3 }}3 + {\log _{\sqrt 3 }}100$ by using multiplication.
And now $100 = {10^2}$ then we have ${\log _{\sqrt 3 }}300 = {\log _{\sqrt 3 }}3 + {\log _{\sqrt 3 }}{10^2}$
Now by use of the property $\log {x^m} = m\log x$, then we get ${\log _{\sqrt 3 }}300 = {\log _{\sqrt 3 }}3 + 2({\log _{\sqrt 3 }}10)$
Let $10 = 2 \times 5,3 = {\sqrt 3 ^2}$ (general algebraic expansion) substituting these values we get ${\log _{\sqrt 3 }}300 = {\log _{\sqrt 3 }}3 + 2({\log _{\sqrt 3 }}10) \Rightarrow {\log _{\sqrt 3 }}{\sqrt 3 ^2} + 2({\log _{\sqrt 3 }}2 \times 5)$
Now using the properties $\log {x^m} = m\log x$ and $\log (ab) = \log a + \log b$ then we get ${\log _{\sqrt 3 }}300 = 2{\log _{\sqrt 3 }}\sqrt 3 + 2({\log _{\sqrt 3 }}2 + {\log _{\sqrt 3 }}5)$
Since in the logarithm, same base value and functions get one which is ${\log _a}a = 1$ then we get ${\log _{\sqrt 3 }}300 = 2 + 2({\log _{\sqrt 3 }}2 + {\log _{\sqrt 3 }}5)$ where ${\log _{\sqrt 3 }}\sqrt 3 = 1$
From the given that we have ${\log _{\sqrt 3 }}5 = a$ and ${\log _{\sqrt 3 }}2 = b$ substituting these values we get ${\log _{\sqrt 3 }}300 = 2 + 2({\log _{\sqrt 3 }}2 + {\log _{\sqrt 3 }}5) \Rightarrow 2 + 2(a + b)$
Thus, we get ${\log _{\sqrt 3 }}300 = 2(a + b + 1)$
Therefore, the option $B)2(a + b + 1)$ is correct.

Note: The other more general properties of the logarithm are
 ${\log _{{a^n}}}b = \dfrac{1}{n}{\log _a}b$ where n is any number that we can choose as the base of the log
$\log (\dfrac{a}{b}) = \log a - \log b$
The log rules can be used for the fast exponent calculation using the multiplication operations. The most general base of the log is $e$