How do you evaluate ${\log _4}(128)$?

Hint: First we write $128\;$ in exponential form. $128\;$ should be written in such a way that it is in the powers of $4$. It is easier if we write it in powers of $4$ so that we can cancel the logarithmic base with it. Use the laws of logarithms to remove the exponent and then simplify it.

Formulas Used:
Law of powers of logarithms, if we have the function, $f(x) = {\log _a}({b^c})$. Then we can convert into power form as, $f(x) = c{\log _a}(b)$ .
Law of sum of logarithms, if we have a function, $f(x) = {\log _c}(ab)$. Then we can write it as $f(x) = {\log _c}a + {\log _c}b$.
Law of base power of logarithms, if we have the expression, $f(x) = {\log _{({b^c})}}a$. It can also be written as, $f(x) = \dfrac{1}{c}{\log _b}c$.

Complete step-by-step answer:
The given logarithmic expression is, ${\log _4}(128)$
Firstly we write $128\;$ in exponential form, as the power of $4$.
Since $128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
We can write $128 = 4 \times 4 \times 4 \times 2$
Now, on converting it into exponential form,
$ \Rightarrow 128 = {4^3} \times 2$
Now put it back into a logarithmic expression.
$ \Rightarrow {\log _4}({4^3} \times 2)$
Using the Law of the sum of logarithms,
 If we have a function, $f(x) = {\log _c}(ab)$. Then we can write it as $f(x) = {\log _c}a + {\log _c}b$.
Here,$a = {4^3};b = 2;c = 4$
$ \Rightarrow {\log _4}({4^3} \times 2) = {\log _4}({4^3}) + {\log _4}2$
Now, consider the first term.
By using the law of powers of logarithms,
If we have the function, $f(x) = {\log _a}({b^c})$. Then we can convert into power form as, $f(x) = c{\log _a}(b)$.
Here $a = 4;b = 4;c = 3$
On Substituting,
$ \Rightarrow {\log _4}({4^3}) = 3{\log _4}4$
If the base and the logarithm value is the same, they cancel out to get $1$
$ \Rightarrow {\log _4}4 = 1$
$ \Rightarrow 3{\log _4}4 = 3$
Now, consider the second term, ${\log _4}2$ which can also be written as ${\log _{({2^2})}}2$.
If we have the expression, $f(x) = {\log _{({b^c})}}a$. It can also be written as, $f(x) = \dfrac{1}{c}{\log _b}a$.
Here, $a = 2;b = 2;c = 2$
On substituting,
$ \Rightarrow {\log _{{2^2}}}2 = \dfrac{1}{2}{\log _2}2$
If the base and the logarithm value is the same, they cancel out to get $1$
$ \Rightarrow {\log _2}2 = 1$
$ \Rightarrow \dfrac{1}{2}{\log _2}2 = \dfrac{1}{2}$
Now, putting it all together we get,
$ \Rightarrow 3 + \dfrac{1}{2} = \dfrac{{(6 + 1)}}{2}$
$ \Rightarrow \dfrac{7}{2} = 3.5$

$\therefore {\log _4}128 = \dfrac{7}{2}$

Note:
One should ensure that the base of the given logarithms is the same before evaluating the expression using the laws of the logarithms. While using the subtraction law of logarithms ensure which term is written in the numerator and which in the denominator. Always the first term will be in the numerator whereas the second term in the denominator.