2(10)^35 = 2(5*2)^35 = (4^18)*(5^35)

So, m = 35

(1/5)^m * (1/4)^18 = 1/2(10)^35, then m = ?

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

=(1/5)^m * (1/2)^36 = 1/2(10)^35

(1/5)^m * (1/2)^35 *1/2 = 1/2(10)^35

now to bring 10 in the deniminator we need to multiply 5*2 no since 2 is raised to the power 35, so m=35 for this to be possible
so m=35

i.e. (1/5)^35 * (1/2)^35 *1/2
=1/(5*2)^35 *1/2

=1/2(10)^35

ans shoulld be D

Hello,

Thank you all for responding. I'll review the proposed solution and post any additional questions.

/Regards.