What is the value of $$\sqrt{\frac{a}{b}}$$, If $$\log_{4}\log_{4}4^{a-b}=2\log_{4}(\sqrt{a}-\sqrt{b})+1$$

$$\sqrt{\frac{a}{b}}$$, If $$\log_{4}\log_{4}4^{a-b}=2\log_{4}(\sqrt{a}-\sqrt{b})+\log_{4}{4}$$

i.e. $$\log_{4}\log_{4}4^{a-b}=\log_{4}((\sqrt{a}-\sqrt{b})^2)*4$$

i.e. $$\log_{4}4^{a-b}=((\sqrt{a}-\sqrt{b})^2)*4$$

i.e. (a-b)*$$\log_{4}4=((\sqrt{a}-\sqrt{b})^2)*4$$

i.e. a-b = 4a+4b-8$$\sqrt{ab}$$

i.e. 3a + 5b - 8$$\sqrt{ab}$$ = 0

i.e. $$3\sqrt\frac{a}{b}^2$$ - 8$$\sqrt\frac{a}{b}$$+5 = 0

put $$\sqrt\frac{a}{b}$$ = t

therefore 3$$t^2$$ - 8t + 5 = 0 

solving we get t = 1 or t = $$\frac{5}{3}$$

i.e. $$\sqrt\frac{a}{b}$$ = 1 or $$\frac{5}{3}$$

but if $$\sqrt\frac{a}{b}$$ = 1 then a=b then $$\log_{4}(\sqrt{a}-\sqrt{b})$$ will become indefinite

Therefore  $$\sqrt\frac{a}{b}$$ = $$\frac{5}{3}$$

Therefore our answer is option 'C'