How do you evaluate: ${\log _{64}}8$ ?

Hint: The logarithmic number is converted to the exponential number. The exponential number is defined as the number of times the number is multiplied by itself. The logarithmic number has a base and we have to find the value of base by conversion. In the given problem, we have to solve for the value of a logarithmic function. So, we will convert it to exponential form and find the value of the variable by comparing the powers.

Complete step by step answer:
The given number is in the form of a logarithmic number and we have to convert it into exponential form. The equation is in the form ${\log _x}y = b$ . To convert it into exponential form it is written as $y = {x^b}$, where x is the base of the log function.

Let us assume ${\log _{64}}8 = t$.Consider the given question ${\log _{64}}8 = t$, when we compare to the general form ${\log _x}y = b$, we get that y is $8$ and x is $64$.
Therefore, it is written as ${64^t} = 8$ in exponential form.
The number $64$ is factored as:
$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2^6}$

Also, the number $8$ can be factored as:
$8 = 2 \times 2 \times 2 = {2^3}$
We know that ${\left( {{2^3}} \right)^2} = {2^6}$. So, we have, ${8^2} = 64$.
Therefore, the number $64$ is written as ${8^2}$.
The above equation is written as:
$ \Rightarrow {64^{\left( {\dfrac{1}{2}} \right)}} = 8$
Comparing this with the equation ${64^t} = 8$, we get,
$\therefore t = \dfrac{1}{2}$

Therefore, the value of the logarithmic function ${\log _{64}}8 = \dfrac{1}{2}$.

Note:To solve the logarithmic equation we need to convert the equation to the exponential form and by using the concept of factorisation we can determine the value of base of the log. The exponential form of a number is defined as the number of times the number is multiplied by itself. The general form of logarithmic equation is ${\log _x}y = b$ and it is converted to exponential form as $y = {x^b}$. Hence, we obtain the result or solution for the equation.