How do you simplify \[{\log _4}16\]?

Hint: Here we will use the basic logarithmic properties to solve the question. First we will rewrite 16 in terms of square of a number. Then we will use the logarithmic property to simplify the terms. We will again use the logarithmic property to find the value of the obtained log function and get the required answer.

Complete step by step solution:
The given function is \[{\log _4}16\].
We know that the number 16 is the square of the number 4 i.e. \[{4^2} = 16\]. Therefore rewriting 16 in terms of the square, we get
\[{\log _4}16 = {\log _4}{4^2}\]
Now we will use the logarithmic property to solve it further.
We know that \[\log {a^b} = b\log a\]. Therefore by using this property in the above equation, we get
\[ \Rightarrow {\log _4}16 = 2{\log _4}4\]
We know by the properties of the log function that the value of \[{\log _a}a = 1\]. So the value of \[{\log _4}4 = 1\]. Therefore by putting this value in the above equation, we get
\[ \Rightarrow {\log _4}16 = 2\left( 1 \right)\]
\[ \Rightarrow {\log _4}16 = 2\]

Hence the value of \[{\log _4}16\] is equal to 2.

Note:
Here in this question we have to modify the number in the given equation according to the given base of log function. We need to keep in mind that the value inside the log function should never be zero or negative; it should always be greater than zero. We should also remember that the value of the \[\log 10\] is equal to 1. The exponential function is the inverse of logarithmic function.
We should also know the basic properties of the log functions. Some of the basic properties are listed below:
\[\begin{array}{l}\log a + \log b = \log ab\\\log {a^b} = b\log a\\\log a - \log b = \log \dfrac{a}{b}\\{\log _a}b = \dfrac{{\log b}}{{\log a}}\end{array}\]