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**Hint:**To solve this question, we will first convert $0.25$ into its fraction form. By doing this we will get the value which is in relation with the base. Then, after applying some properties of logarithms, we will solve this question.

**Complete step-by-step solution:**

We have ${\log _2}(0.25)$.

First, we will convert $0.25$into fraction form.

$0.25 = \dfrac{{25}}{{100}} = \dfrac{1}{4}$

Therefore, we can write

${\log _2}(0.25) = {\log _2}\left( {\dfrac{1}{4}} \right)$

4 can be written as ${2^2}$

\[

{\log _2}(0.25) \\

= {\log _2}\left( {\dfrac{1}{4}} \right) \\

= {\log _2}\left( {\dfrac{1}{{{2^2}}}} \right) \\

= {\log _2}\left( {{2^{ - 2}}} \right) \\

\]

Here, we have used the property $\dfrac{1}{{{x^n}}} = {x^{ - n}}$

Now, we will use the power rule for logarithms. This rule states that logarithm of a positive number a to the power b is equal to the product of b and log of a which means that \[\log {a^b} = b\log a\].

Therefore in our case,

\[{\log _2}{2^{ - 2}} = - 2{\log _2}2\]

Now we will apply the rule that the logarithm of a number with the same base is equal to one which means that \[{\log _a}a = 1\].

Therefore in our case, \[{\log _2}2 = 1\]

\[ \Rightarrow - 2{\log _2}2 = - 2 \times 1 = - 2\]

Thus, ${\log _2}(0.25) = - 2$.

Hence, our final answer is $ - 2$.

**Note:**This problem can be solved by another method also. In this method, we can use the change of base rule.

${\log _a}x = \dfrac{{{{\log }_b}x}}{{{{\log }_b}a}}$

This rule can be applied if both the numbers $a$and $b$ are greater than zero and not equal to one.

Here, we have ${\log _2}(0.25)$.

So, if we compare this with ${\log _a}x$, we have $a = 2$ and $x = 0.25$.

Here, we will change the base from 2 to 10. Therefore, $b = 2$.

Thus, we can write

\[{\log _2}0.25 = \dfrac{{{{\log }_{10}}0.25}}{{{{\log }_{10}}2}}\]

we will convert $0.25$into fraction form.

$0.25 = \dfrac{{25}}{{100}} = \dfrac{1}{4}$

Therefore, we have

\[{\log _2}0.25 = \dfrac{{{{\log }_{10}}\left( {\dfrac{1}{4}} \right)}}{{{{\log }_{10}}2}} = \dfrac{{{{\log }_{10}}\left( {\dfrac{1}{{{2^2}}}} \right)}}{{{{\log }_{10}}2}} = \dfrac{{{{\log }_{10}}\left( {{2^{ - 2}}} \right)}}{{{{\log }_{10}}2}}\]

Now, we will use the power rule for logarithms. This rule states that logarithm of a positive number a to the power b is equal to the product of b and log of a which means that \[\log {a^b} = b\log a\].

Therefore in our case,

\[{\log _2}{2^{ - 2}} = - 2{\log _2}2\]

\[{\log _2}0.25 = \dfrac{{{{\log }_{10}}\left( {\dfrac{1}{4}} \right)}}{{{{\log }_{10}}2}} = \dfrac{{{{\log }_{10}}\left( {\dfrac{1}{{{2^2}}}} \right)}}{{{{\log }_{10}}2}} = \dfrac{{{{\log }_{10}}\left( {{2^{ - 2}}} \right)}}{{{{\log }_{10}}2}} = \dfrac{{ - 2{{\log }_{10}}2}}{{{{\log }_{10}}2}} = - 2\]

Thus, we can solve this problem by using two different methods.

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