How do you evaluate \[{\log _{16}}1?\]

Hint: The logarithm base $b$ of $x$ , where both $b$ and $x$ should be positive. Everything inside the log must be raised to the same exponent in order to pull the exponent out. To solve this equation, we have to generally use the theorem of Logarithm which is given by. When positive real numbers and $b$ does not equal to $1$ , then ${\log _b}(x) = y$ is equal to $x = {b^y}$.

Complete step by step solution:
We know that this equation can be easily solved by using property of logarithm for the given question,
we have to evaluate \[{\log _{16}}1\]
Let us assume \[{\log _{16}}1 = x\]
By applying the theorem which we know that
${\log _b}x = y$ is equal to $x = {b^y}$
On comparing and putting the respective value we get
$1 = {(16)^x}$
Since, we know that we can write
${(2)^0} = 1$
Also we know that $16 = {2^4}$
Thus on simplifying, we get
${(2)^0} = {\left( {{2^4}} \right)^x}$
We know the property ${({a^x})^y} = {a^{xy}}$ .
Use the property in above equation and we get
$ \Rightarrow {(2)^0} = {(2)^{4x}}$
After, comparing from both side, we get
$0 = 4x$
$ \Rightarrow x = 0$
So, the value of \[{\log _{16}}1\] is $0$.

Note:
To solve the above equation we must have the basic knowledge of the basic theorem of logarithm and some basic mathematics formulae and understanding. These properties of logarithm are very useful in solving very complex equations in mathematics.