Given $\log 6\ and\ \log 8$, then the only logarithm that cannot be obtained without using the table is,
A. $\log 64$
B. $\log 21$
C. $\log \dfrac{8}{3}$
D. $\log 9$

Hint: We will be using the concepts of logarithm functions to solve the problem. We will start by using the properties of logarithm that $\log a\times b=\log a+\log b$ then we will use this property to express the given options in terms of $\log 6\ and\ \log 8$ in the process we will find the option which cannot be expressed in the form of $\log 6\ and\ \log 8$ so we will use the approach of option checking to further make the process of finding a solution easy.

Complete step-by-step answer:
Now, we have been given $\log 6\ and\ \log 8$ and we have to find the logarithm given in options which cannot be found by using the $\log 6\ and\ \log 8$.
So, in option (A) we have $\log 64$.
Now, we know that ${{\log }_{a}}{{b}^{m}}=m{{\log }_{a}}b$.
Therefore,
$\begin{align}
  & \log 64=\log {{8}^{2}} \\
 & =2\log 8 \\
\end{align}$
Since, we know $\log 8$. Therefore, $\log 64$ can be found.
Now, in option (B) we have $\log 21$.
Now, we know the logarithmic identity that,
$\log a\times b=\log a+\log b$
So, we have
$\begin{align}
  & \log 21=\log 3\times 7 \\
 & =\log 3+\log 7 \\
\end{align}$
Now, $\log 3$ can be found. We have been given,
$\begin{align}
  & \log 6=\log 2\times 3=\log 2+\log 3 \\
 & \log 6=\log 2+\log 3 \\
\end{align}$
Now, we will multiply and divide by 3 with $\log 2$.
$\begin{align}
  & \log 6=\dfrac{3\log 2}{3}+\log 3 \\
 & \log 6=\dfrac{\log {{2}^{3}}}{3}+\log 3 \\
 & \log 6=\dfrac{\log 8}{3}+\log 3..........\left( 1 \right) \\
\end{align}$
So, we can find the value of $\log \left( 3 \right)$ from (1) but $\log \left( 7 \right)$ cannot be found using a similar method.
Now similarly we have option (C ) as $\log \left( \dfrac{8}{3} \right)$
Now we know the logarithmic identity that
$\log \left( \dfrac{a}{b} \right)=\log \left( a \right)-\log \left( b \right)$
Therefore the option (c ) can be written as
 $\log \left( \dfrac{8}{3} \right)=\log \left( 8 \right)-\log \left( 3 \right)$
Now we know the value of $\log \left( 8 \right)$ from the data given to us and we have the value of $\log \left( 3 \right)$ from the equation (1) so we can determine the value of option ( C) .
Now in option (D) we have $\log 9$
We will now use the logarithmic identity that $\log \left( {{a}^{m}} \right)=m\log \left( a \right)$
$\begin{align}
  & \log \left( 9 \right)=\log \left( {{3}^{2}} \right) \\
 & \Rightarrow \log \left( {{3}^{2}} \right)=2\log \left( 3 \right) \\
\end{align}$
Now we know the value of $\log \left( 3 \right)$from equation (1) so we can find the value of the option (D) also.
Therefore, option (B) can’t be obtained without using logarithmic table.
Hence, the answer of the question is option (B).

Note: To solve these type of questions it is important to remember logarithmic identities like,
$\begin{align}
  & \log \left( a\times b \right)=\log a+\log b \\
 & \log {{b}^{m}}=m\log b \\
 & \log \dfrac{a}{b}=\log a-\log b \\
\end{align}$
It is important to note how we have used the value of $\log 6\ and\ \log 8$ to find the value of $\log \left( 3 \right)$ by using the logarithmic identities mentioned above this trick is crucial in solving the options (C ) and (D).