How do you evaluate ${\log _6}(2)?$

Hint: WE can see that we have been given a logarithmic function in the above question, where the base of the function is $6$ and the term with logarithm is $2$ . So in this question, we will use the properties and formula of the logarithm to simplify and calculate the given expression. The main property that we will use here is the change of base property of the logarithm functions.

Formula used:
${\log _a}b = \dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}$
Product of the logarithmic formula is
$\log ab = \log a + \log b$ .

Complete step-by-step answer:
Here we have the given function ${\log _6}(2)$ .
We should know that according to base property, the base of the logarithm function can be changed to base $e$ , which is the natural log, but then the function can be expressed as the division of logarithmic base $e$ of argument divided by the logarithm base $e$ of the base given in the original question.
Now by comparing the function with the base formula, we have
 $a = 6,b = 2$
So by applying the base formula we can write:
$\dfrac{{{{\log }_e}2}}{{{{\log }_e}6}}$
Now we can see that in the denominator we can write $6$ as the product of $2 \times 3$
So by applying the product formula of the logarithm, we can write the function as:
 $\dfrac{{{{\log }_e}2}}{{\log 2 \times \log 3}}$
We know the value that:
$\log 2 = 0.693,\log 3 = 1.098$
Now we will substitute the values in the equation we have:
$\dfrac{{0.693}}{{0.693 + 1.098}}$
We will now simplify the given value:
$ \Rightarrow {\log _6}2 = \dfrac{{0.693}}{{1.791}}$
$ \Rightarrow {\log _6}2 = 0.386853$
Hence the required answer is $0.386853$ (approx.).

Note: Before solving this kind of question, we must have prior knowledge of logarithm functions and their formulas. We should note that if no base is given in the function, that it is assumed to have the base $10$ . We should know some of the basic logarithmic formulas to solve questions such as:
The Power rule of the logarithm is
${\log _a}({x^n}) = n{\log _a}x$
Quotient rule:
${\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y$
Zero Rule:
${\log _a}1 = 0$
${\log _a}a = 1$