I can't seem to figure out the following problem. I think I'm just blanking on a rule that I should be applying...
[(1/5)^m] * [(1/4)^18] = 1/[2 * (10)^35]
a) 17
b) 18
c) 34
d) 35
e) 36
Thanks...
Help with exponent problem...
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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.
[(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.
- indiantiger
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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]
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.
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=> 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
=> 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
- eaakbari
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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
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