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

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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?  [#permalink]

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New post 05 Apr 2016, 22:31
Attached is a visual that should help.
Attachments

Screen Shot 2016-04-05 at 10.29.41 PM.png
Screen Shot 2016-04-05 at 10.29.41 PM.png [ 123.7 KiB | Viewed 1690 times ]


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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?  [#permalink]

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New post 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 = ?  [#permalink]

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New post 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 = ?  [#permalink]

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New post 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 = ?  [#permalink]

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New post 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

Answer D

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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?  [#permalink]

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New post 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 = ?  [#permalink]

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New post 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 = ?  [#permalink]

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New post 15 May 2017, 17:32
1
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 re-expressed 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 = ?  [#permalink]

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New post 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 = ?  [#permalink]

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New post 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

\((\frac{1}{5})^m * (\frac{1}{4})^{18} = \frac{1}{2*(10)^{35}}\)

\((5)^{-m} * (2)^{-2*18} = 2^{-36}*(5)^{-35}\)

\(5^{-m} * 2^{-36}-2^{-36}*(5)^{-35}=0\)

\(2^{-36}(5^{-m}-5^{-35})=0\)

\(5^{-m}=5^{-35}\)

\(-m=-35\)

\(m=35\)
Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ? &nbs [#permalink] 21 Aug 2018, 10:16

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