Confusing Order

$\Large 2^{\left(\frac{1}{500}\right){3000}} = 2^{\frac{3000}{500}} = 2^{6}$

$\Large 3^{\left(\frac{1}{300}\right){3000}} = 3^{\frac{3000}{300}} = 3^{10}$

$\Large 4^{\left(\frac{1}{100}\right){3000}} = 4^{\frac{3000}{100}} = 4^{30}=(4^6)^5$

$\Large 5^{\left(\frac{1}{200}\right){3000}} = 5^{\frac{3000}{200}} = 5^{15} =(5^3)^5$

We clearly know that $\Large 4^6 > 125 \Rightarrow 4^{30} >5^{15}$

Thus we have the inequality:

$\Large{ 2^6 < 3^{10} < 5^{15} < 4^{30} \\ \Rightarrow 2^{\frac{3000}{500}} < 3^{\frac{3000}{300}} <5^{\frac{3000}{200}} <4^{\frac{3000}{100}} \\ \Rightarrow 2^{\left(\frac{1}{500}\right){3000}} < 3^{\left(\frac{1}{300}\right){3000}} < 5^{\left(\frac{1}{200}\right){3000}} < 4^{\left(\frac{1}{100}\right){3000}} \\ \Rightarrow \boxed{2^{\frac{1}{500}} < 3^{\frac{1}{300}} < 5^{\frac{1}{200}} < 4^{\frac{1}{100}}}}$

Thus the decreasing order is $d,c,b,a$ .

if we take log in all options it wont effect the order so

a. $log 2^{\large\frac{1}{500}}$ = $\frac{1}{500} log 2$

b. $log \large 3^{\frac{1}{300}}$ = $\frac{1}{300} log 3$

c. $log \large 5^{\frac{1}{200}}$ = $\frac{1}{200} log 5$

d. $log \large 4^{\frac{1}{100}}$ = $\frac{1}{100} log 4$

Now as log 2 $\approx$ log 3 $\approx$ log 4 $\approx$ log 5 = log n

a = $\frac{1}{500} log (n)$

b = $\frac{1}{300} log (n)$

c = $\frac{1}{200} log (n)$

d = $\frac{1}{100} log (n)$

$\large\boxed{ \Longrightarrow d > c > b > a}$