I'm guessing the question is to solve for m?

[(1/5)^m] * [(1/4)^18] = 1/[2 * (10)^35]

[(1/5^m)] * [(1/4^18)] = 1/[2 * (2)^35* (5)^35]

[(1/5^m)] * [(1/2^36)] = 1/[(2)^36*(5)^35]

1/[(2)^36*(5)^m] = 1/[(2)^36*(5)^35]

m=35

Let me know if you need help with any of the workings.

Given =

(1/5)^m * (1/4)^18 = 1/[2*(10)^35]

lets simplify LHS first

=(1/5)^m * (1/2) ^2*18
=(1/5)^m * (1/2)^36

Now lets simplify RHS
= 1/[2(5*2)^35]
= 1/2*(1/2)^35*(1/5)^35
=(1/2)^36*(1/5)^35


LHS = RHS

(1/5)^m*(1/2)^36 = (1/2)^36 *(1/5)^35

the only way LHS = RHS is for m = 35

(D)

Please feel free to correct me if I have done anything wrong.

=> 1/5^m * 1/4^18 = 1/2 *10^35

=> 2 *10^35 = 5^m* 4^18

=> [2 *10^35]/ 4^18 = 5^m

=> [2 *10^35]/ 2^36 = 5^m

=> [10^35]/ 2^35 = 5^m

=> [10/ 2]^35 = 5^m

=> 5^35 = 5^m

=> m = 35

When you do look at the question, you should realize that theres 5 , 4 and 10. 5*2=10. So its preferable to make 4 in terms of two.Lets proceed

5^-m * 4^-18 = 1/2 * 10^-35

5^-m * 2^-36 = 1/2 * 10^-35

5^-m * 2^-35 = 10^-35

Which obviously implies m = 35

So D