Hi all, I just came across following question:
(1/5)^m (1/4)^18 = 1 / [2(10)^35]
Question: what is the value of m?
(A) 17
(B) 18
(C) 34
(D) 35
(E) 22
I have no idea how to do this and would appreciate your help.
Exponents problem
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- rohit_gmat
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hi.. i guess this is from the 700+ list...schreze wrote:Hi all, I just came across following question:
(1/5)^m (1/4)^18 = 1 / [2(10)^35]
Question: what is the value of m?
(A) 17
(B) 18
(C) 34
(D) 35
(E) 22
I have no idea how to do this and would appreciate your help.
here is how to do it..
1/ [ 2 x (10^35] = (1/5)^m x (1/2)^(18x2) = [1/(5^m)] x [1/ (2^36)]
take 5 x 2 = 10 and 5^a x 2^a = 10^a
so we have
1/ [ 2 x (10^35] = 1/ [ (5^m) x (2^35) x 2]
which suits the original eq.. so we proceed as
1/ [ 2 x (10^35] = 1/ [ 2 x (5 x 2)^35]
1/ [ 2 x (10^35] = 1/ [ 2 x (10^35]
so m has to be 35 in order for the eq to be true...
Answer is D
- cans
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as 5 is prime number, power of 5 on lhs and rhs should be same.
LHS: -m (5 is raised to power -m)
RHS: -35 (5 is raised to power -35)
Thus -m=-35 or m=35
IMO D
LHS: -m (5 is raised to power -m)
RHS: -35 (5 is raised to power -35)
Thus -m=-35 or m=35
IMO D
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- mehrasa
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we consider the denominators:
5^m * 2^36 = 2*10^35
at this stage we take out 2^m from 2^36 in order to make the base 10 ..... so:
5^m * 2^m * 2^(36-m) = 10^m * 2^(36-m)==> m=35
hope it helps
5^m * 2^36 = 2*10^35
at this stage we take out 2^m from 2^36 in order to make the base 10 ..... so:
5^m * 2^m * 2^(36-m) = 10^m * 2^(36-m)==> m=35
hope it helps