Explanation
Let $$\log { 6 } =x$$
$$\Rightarrow \log { 2 } +\log { 3 } =x$$ ....... $$(i)$$
Let $$\log { 8 } =y$$
$$\Rightarrow 3\log { 2 } =y$$
$$\Rightarrow \log 2=\dfrac{y}{3}$$ ...... $$(ii)$$
$$\therefore x=\log { 3 } +\dfrac { y }{ 3 } $$ ...... From $$(i)$$ and $$(ii)$$
$$\Rightarrow \log { 3 } =x-\dfrac { y }{ 3 } $$
A.
$$\log { 64 } =\log { { 2 }^{ 6 } }=6\log 2=6\dfrac{y}{3} $$
B.
$$\log { 21 } =\log { 3 } +\log {7} $$
Thus, $$\log 21$$ cannot be found from given values as we need the value of $$\log { 7 } $$.
C.
$$\log \dfrac{8}{3}=\log 8-\log 3=3\log 2-\log 3$$
D.
$$\log 9=\log 3^{2}=2\log 3$$
Hence, only $$\log 21$$ can't be obtained from given data.
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