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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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05 Apr 2016, 22:31
Attached is a visual that should help.
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Screen Shot 20160405 at 10.29.41 PM.png [ 123.7 KiB  Viewed 1962 times ]
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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24 May 2016, 18:53
Can someone please explain where the 1/2^36 is coming from?



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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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24 May 2016, 20:53
sahmedmartinez wrote: Can someone please explain where the 1/2^36 is coming from? (1/5)^m * (1/4)^18 = 1 / (2 * (10)^35) = (1/5)^m * (1/2^2)^18 = 1/( 2* (2 * 5)^35) Two formulas are used here: 1. (a^m)^n = a^(mn) 2. a^m * a^n = a^(m + n) =(1/5)^m * (1/2^36) = 1/(2^36 * 5^35)



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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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16 Dec 2016, 04:02
Tessa8 wrote: If \(([1][/5]^m [1][/4]^(18) = [1][/2(10)^(35)]\) , then m = ?
a) 17 b) 18 c) 34 d) 35 e) 36
screenshot of question attached (1/4)^18==[(1/2)^2]^18==(1/2)^36 also (1/10)^35==[1/(2*5)]^35 putting above both values in equation we get (1/5)^m(1/2)^36=1/2*[1/(2*5)]^35 (1/5)^m(1/2)^36=[1/2]^36*(1/5)^35 cancelling 1/2^36 both sides (1/5)^m=(1/5)^35 thus equating powers we get m=35 Ans D



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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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11 May 2017, 05:55
(1/5)^m * (1/4)^18 = 1/(2*(10)^35) (1/5)^m * (1/2)^36 = 1/(2*(2*5)^35) (1/5)^m * (1/2)^36 = 1/(2^36) * (1/5^35) For the two sides of the equation to be true, the powers of each base should be the same. => m = 35
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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11 May 2017, 06:40
domu904 wrote: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
A. 17 B. 18 C. 34 D. 35 E. 36 (1/5)^m * (1/4)^18 = 1/(2*(10)^35) > \(\frac{(1)}{(5^m*2^3^6)}=\frac{(1)}{(2^3^6*5^3^5)}\) m = 35
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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12 May 2017, 15:26
Pretty straight forward. M=35. I fel like asking for what r is would be more challenging. As it would require to realise that 2 is to the power of 36, so r=18. Great question though! So this is a low 600 level question?



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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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13 May 2017, 04:04
vmelgargalan wrote: Pretty straight forward. M=35. I fel like asking for what r is would be more challenging. As it would require to realise that 2 is to the power of 36, so r=18. Great question though! So this is a low 600 level question? Yes it is a pretty straight forward question and hence below 600 level question... You can just solve the question in mind. Will not take more than 20 seconds..
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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15 May 2017, 17:32
domu904 wrote: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
A. 17 B. 18 C. 34 D. 35 E. 36 The negative exponent rule states: (1/a)^b can be reexpressed as a^(b). Let’s flip all of the fractions using the negative exponent rule: 5^m x 4^18 = 2^1 x 10^35 5^m x 2^36 = 2^1 x 2^35 x 5^35 5^m x 2^36 = 2^36 x 5^35 5^m = 5^35 m = 35 Answer: D
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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21 Aug 2017, 09:14
I feel the question is pretty straight forward but ans was incorrect in many of the docs floating.
My take was
Left hand side denominator : 5^M Right hand side denominator: 10^35 implies 5^35 * 2^35 So M=35



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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?
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21 Aug 2018, 10:16
domu904 wrote: If \((\frac{1}{5})^m * (\frac{1}{4})^{18} = \frac{1}{2*(10)^{35}}\), then m = ?
A. 17 B. 18 C. 34 D. 35 E. 36 OA:D\((\frac{1}{5})^m * (\frac{1}{4})^{18} = \frac{1}{2*(10)^{35}}\) \((5)^{m} * (2)^{2*18} = 2^{36}*(5)^{35}\) \(5^{m} * 2^{36}2^{36}*(5)^{35}=0\) \(2^{36}(5^{m}5^{35})=0\) \(5^{m}=5^{35}\) \(m=35\) \(m=35\)



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Re: 1) if (1/5)^m * )1/4)^18 = 1/2(10)^35, then m=? 2) for every
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Re: 1) if (1/5)^m * )1/4)^18 = 1/2(10)^35, then m=? 2) for every
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