Question 59

# $$\frac{log (97-56\sqrt{3})}{log \sqrt{7+4\sqrt{3}}}$$ equals which of the following?

Solution

Simplify the expression a bit to remove the root sign in the denominator

$$\dfrac{\log{97-56\sqrt{3}}}{\dfrac{1}{2}\times (\log{7+4\sqrt{3})}}$$

$$\Rightarrow 2 \times \dfrac{\log{97-56\sqrt{3}}}{\log{7+4\sqrt{3}}}$$

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 $$a + b - (2 \times \sqrt{a} \sqrt{b}) = a+b-2\sqrt{ab}$$

comparing the value of terms under root with the terms in the numerator, we get

$$\sqrt{ab} = 28\sqrt{3}$$ and $$a+b=97$$

From solving this, we get to know that $$a=7$$ and $$b=4\sqrt{3}$$

Thus the expression can be written as $$2 \times2 \times \dfrac{\log{7-4\sqrt{3}}}{\log{7+4\sqrt{3}}}$$

$$\Rightarrow 4 \times \dfrac{\log{7-4\sqrt{3}}}{\log{7+4\sqrt{3}}}$$

Now, let us look at the reciprocal of the term in log in the denominator.

$$\frac{1}{7+4\sqrt{\ 3}} = \frac{1}{7+4\sqrt{\ 3}} \times \dfrac{7-4\sqrt{3}}{7-4\sqrt{3}}$$

$$\Rightarrow \dfrac{7-4\sqrt{3}}{7^{2}-(4\sqrt{3})^{2}}$$

$$\Rightarrow \dfrac{7-4\sqrt{3}}{49-48} = 7-4\sqrt{3}$$

$$\frac{\log\left(7-4\sqrt{\ 3}\right)}{\log\left(7+4\sqrt{\ 3}\right)}=\frac{\log\left(7-4\sqrt{\ 3}\right)}{\log\left(\frac{1}{7-4\sqrt{\ 3}}\right)}=\frac{\log\left(7-4\sqrt{\ 3}\right)}{-\log\left(7-4\sqrt{\ 3}\right)}=-1$$

Thus the value of the expression can be further simplified as

$$\Rightarrow 4 \times (-1) = -4$$

Hence the correct answer is option C