How do you write ${{\log }_{8}}64=2$ in exponential form? \[\]

Hint: We recall exponential from and logarithmic form. We recall if we express $x$ in exponential form with base $b$ and exponent $y$ as ${{b}^{y}}=x$ then we can express the exponent $y$ in logarithmic from as ${{\log }_{b}}x=y$. We can also convert from logarithmic form to exponential form as ${{\log }_{b}}x=y\Rightarrow {{b}^{y}}=x$. We take $b=8,x=64,y=2$ to write in exponential form. \[\]

Complete step-by-step solution:
We know that if we multiply a number $b$ with itself $y$ times and get the product $x$ then we can write in exponential form as
\[b\times b\times b\times ....\left( \text{y times} \right)={{b}^{y}}=x\]
Here $b$ is called base and $y$ is called exponent or power or an index of the base. Here the base and exponent cannot be zero at the same time. We know that the logarithm is the inverse operation to exponentiation. That means the logarithm of a given number $x$ is the exponent to which another fixed number, the base $b$ must be raised, to produce that number$x$, which means if ${{b}^{y}}=x$ then the logarithm denoted as log and calculated as
\[{{\log }_{b}}x=y\]
Here $x$ is called the argument of the logarithm and is always positive. Here the base of the logarithm $b$ has more restrictions which are $b>0,b\ne 1$. We are given an expression in logarithmic from as
\[{{\log }_{8}}64=2\]
We see that here the base is $b=8$ , the argument is $x=64$ and the logarithmic value $y=2$. So we convert it into exponential form with base $b=8$, exponent $y=2$ and result of exponentiation $x=64$ to have;
\[{{8}^{2}}=64\]

Note: We note that we can always convert from logarithmic form to exponential form but it may not be so from exponential form to logarithmic form for example ${{\left( -1 \right)}^{-3}}=-1$ cannot be converted into logarithm form because here we have $b=-1,x=-1$. The logarithm with base 10 is called common logarithm and is written without the base as $\log x$.