Simplify the expression a bit to remove the root sign in the denominator $$dfrac{log{97-56sqrt{3}}}{dfrac{1}{2} imes (log{7+4sqrt{3})}}$$ $$ Rightarrow 2 imes dfrac{log{97-56sqrt{3}}}{log{7+4sqrt{3}}} $$  To move further, let us see the root of the numerator.  Assume the root of the numberator to be $$sqrt{a}-sqrt{b}$$. When we square it, we get $$a + b - (2 imes sqrt{a} sqrt{b}) = a+b-2sqrt{ab} $$ comparing the value of terms under root with the terms in the numerator, we get  $$sqrt{ab} = 28sqrt{3} $$ and $$a+b=97$$  From solving this, we get to know that $$a=7$$ and $$b=4sqrt{3}$$  Thus the expression can be written as $$ 2 imes2 imes dfrac{log{7-4sqrt{3}}}{log{7+4sqrt{3}}} $$ $$ Rightarrow 4 imes dfrac{log{7-4sqrt{3}}}{log{7+4sqrt{3}}} $$  Now, let us look at the reciprocal of the term in log in the denominator.  $$ dfrac{1}{log{7+4sqrt{3}}} = dfrac{1}{log{7+4sqrt{3}}} imes dfrac{7-4sqrt{3}}{7-4sqrt{3}} $$ $$ Rightarrow dfrac{7-4sqrt{3}}{7^{2}-(4sqrt{3})^{2}} $$
$$ Rightarrow dfrac{7-4sqrt{3}}{49-48} = 7-4sqrt{3}$$
Thus the value of the expression can be further simplified as  $$ 4 imes dfrac{(-1) imes (7+4sqrt{3})}{7+4sqrt{3}} $$ $$ Rightarrow 4 imes (-1) = -4 $$
Hence the correct answer is option C