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Shortcut approach: \((\frac{1}{5})^m * (\frac{1}{4})^{18} = \frac{1}{2*10^{35}}\) --> \(\frac{1}{5^m}* (\frac{1}{4})^{18} = \frac{1}{2*2^{35}*5^{35}}\) --> as there are only integers in the answer choices then we can concentrate only on the power of 5: they should be equal on both sides --> \(m=35\).

Since you are asked to find the power of 5 i.e. m, you should be looking for 5 on the right hand side of the equation raised to some power. There is 10 on the denominator which is 2 * 5 raised to the power 35. The rest of the solution is rearranging to compare the powers of 5 on both sides. _________________

Yogi Bhajan: If you want to learn a thing, read that; if you want to know a thing, write that; if you want to master a thing, teach that. This message transmitted on 100% recycled electrons.

Last edited by hb on 10 Aug 2013, 20:07, edited 1 time in total.

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

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23 Aug 2014, 23:26

Just need to break the equations a bit: 1. (1/5)^m (1/4)^18 = (1/5)^m (1/2)^36 = so now we have 36 powers of 1/2 and need to find for 5 , what we need is the relation between this equation and other so we will try to sync them up. 2. 1/(2(10)^35) = 1/(2(2*5)^35 = so now we have 36 powers of 2 and 35 powers of 5 Finally what we need is how many powers of 5?? its 35 , OA:D. Hope its clear _________________

Comparing left and right side (1/2)^36 is same on both sides . For equation to be true (1/5)^35 should equal to (1/5)^m which is possible when m=35 _________________

I had this question on the GMATPrep 1 and I can't figure it out, could someone please help

(1/5)^m (1/4)^18 = 1/(2(10)^35)

m= 35

This was a problem solving question but i do not understand how they got m=35!

Thanks for your help!!

Dear annaleroy, My friend, perhaps you are new to the GC forum, but you have committed a classic no-no. You see, this very math question has been posted many times before, for example: if-1-5-m-1-4-18-frac-127321.html 1-5-m-1-4-18-1-2-10-35-what-is-m-25136.html if-1-5-m-1-4-18-1-2-10-35-then-m-121477.html if-1-5-m-1-4-18-1-2-10-35-then-m-92796.html Rather than starting a brand new thread, users are supposed to find the already existing threads that discuss the same question. It may be that you will find your answer already posted there, and if you don't, you can ask your question as part of that thread, thus enhancing that discussion, rather than starting a new and separate discussion.

In all likelihood, Bunuel will merge this thread into one of these pre-existing threads.

I had this question on the GMATPrep 1 and I can't figure it out, could someone please help

(1/5)^m (1/4)^18 = 1/(2(10)^35)

m= 35

This was a problem solving question but i do not understand how they got m=35!

Thanks for your help!!

Dear annaleroy, My friend, perhaps you are new to the GC forum, but you have committed a classic no-no. You see, this very math question has been posted many times before, for example: if-1-5-m-1-4-18-frac-127321.html 1-5-m-1-4-18-1-2-10-35-what-is-m-25136.html if-1-5-m-1-4-18-1-2-10-35-then-m-121477.html if-1-5-m-1-4-18-1-2-10-35-then-m-92796.html Rather than starting a brand new thread, users are supposed to find the already existing threads that discuss the same question. It may be that you will find your answer already posted there, and if you don't, you can ask your question as part of that thread, thus enhancing that discussion, rather than starting a new and separate discussion.

In all likelihood, Bunuel will merge this thread into one of these pre-existing threads.

Does all this make sense? Mike

Hi Mike,

Yes! Thank you very much! Won't be doing this going forward!! Thank you!

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

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04 Mar 2015, 06:17

devilbart wrote:

annaleroy wrote:

Hey!

I had this question on the GMATPrep 1 and I can't figure it out, could someone please help

(1/5)^m (1/4)^18 = 1/(2(10)^35)

m= 35

This was a problem solving question but i do not understand how they got m=35!

Thanks for your help!!

Pretty Straight forward. forget about the LHS and re arrange the R.H.S to match the LHS fractions

RHS:

1/(2(2*5)^35)

1/(2^36*5^35)

1/(4^18 * 5^35) ...... [ 2^36 is same as saying (2^2)^18]

break the denominator as per the LHS

(1/4)^18 * (1/5)^35

and Whalaa you have Mr. m

I got a few stupid questions:

1) What is RHS / LHS? 2) How come you can go from 4^18 --> 2^36 but not from, say 5^4 --> 10^2? When does the "divide by 2 and multiply by 2 rule apply"?

3) 1/5^m - How do I get rid of the ^m in the numerator? I mean, isnt 1/5^m \(1^m/5^m?\)

I had this question on the GMATPrep 1 and I can't figure it out, could someone please help

(1/5)^m (1/4)^18 = 1/(2(10)^35)

m= 35

This was a problem solving question but i do not understand how they got m=35!

Thanks for your help!!

Pretty Straight forward. forget about the LHS and re arrange the R.H.S to match the LHS fractions

RHS:

1/(2(2*5)^35)

1/(2^36*5^35)

1/(4^18 * 5^35) ...... [ 2^36 is same as saying (2^2)^18]

break the denominator as per the LHS

(1/4)^18 * (1/5)^35

and Whalaa you have Mr. m

I got a few stupid questions: no doubts are stupid and its always better to clear your basic doubt then to remain with them

hi erik,

1) What is RHS / LHS? RHS is right hand side of equality sign and LHS is left hand side ... 2) How come you can go from 4^18 --> 2^36 but not from, say 5^4 --> 10^2? When does the "divide by 2 and multiply by 2 rule apply"? 4^18=(2*2)^18=\((2^2)\)^18=2^(2*18)=2^36..... 5^4=(5^2)^2=25^2

3) 1/5^m - How do I get rid of the ^m in the numerator? I mean, isnt 1/5^m \(1^m/5^m?\).. 1^m=1 for all values of 1.... if we divide or multiply 1 any number of times, the result will be 1 so m can be removed from power.. _________________

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