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

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If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ? [#permalink] New post 09 Feb 2012, 13:17
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If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?

A. 17
B. 18
C. 34
D. 35
E. 36
[Reveal] Spoiler: OA
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Re: If (1/5)^{m} (1/4)^{18} = \frac{1}{2(10)^{35} [#permalink] New post 09 Feb 2012, 13:24
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If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ?

A. 17
B. 18
C. 34
D. 35
E. 36

Step by step:
\((\frac{1}{5})^m*(\frac{1}{4})^{18}= \frac{1}{2*10^{35}}\) --> \(\frac{1}{5^m}*\frac{1}{2^{36}}= \frac{1}{2*2^{35}*5^{35}}\) --> \(\frac{1}{5^m}*\frac{1}{2^{36}}= \frac{1}{2^{36}*5^{35}}\) --> \(\frac{1}{5^m}= \frac{1}{5^{35}}\) --> \(m=35\).

Answer: D.

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 --> \(m=35\).

Answer: D.
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Re: If (1/5)^{m} (1/4)^{18} = \frac{1}{2(10)^{35} [#permalink] New post 09 Feb 2012, 13:31
Thanks Bunnuel for the quick reply. So does that mean 1^m will always equal to 1? Because I was assuming m could be negative integer.
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Re: If (1/5)^{m} (1/4)^{18} = \frac{1}{2(10)^{35} [#permalink] New post 09 Feb 2012, 13:39
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If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink] New post 02 Apr 2013, 06:58
steliossb wrote:
hey guys

this is the first question i got from the prep program and i am having trouble figuring it out:

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

find M


A) 17
B) 18
C) 34
D) 35
E) 36



any help will greatly be appreciated



4=2^2 => (1/2)^36*5^m=2*10^35

in order to get 10^35 you have to have 2^35*5^35. since we have 2^36, 2 will be left and the answer will be 2*10^35
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Re: I got this question on a Mock GMAT and I hate it so much [#permalink] New post 10 Aug 2013, 18:53
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sokenyou wrote:
(1/5)^m * (1/4)^18 = 1/2(10)^35

What is the value of m?

1. 17
2. 18
3. 34
4. 35
5. 36

I just don't get how to solve it using backsolving or picking numbers or something that is quick....I just hate it... :x


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

\((5^-1)^m * (2^-2)^18 = (2^-1 * 2^-35 * 5^-35)\)

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

\(5^(-m) = * 5^(-35)\)

\(- m = -35\)

\(m = 35\)

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.
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Last edited by hb on 10 Aug 2013, 19:07, edited 1 time in total.
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Re: I got this question on a Mock GMAT and I hate it so much [#permalink] New post 11 Aug 2013, 00:32
In these types of problems we can compare(qualize) the indices of the comparable variables.

a^m = a^n implies m=n

So the only term on the left hand side with five in it is 1/5 ^m
The only term on the right hand side with 5 in it is (1/10)^35

as 1/10 contains only one five so the resultant power on the right hand side is (1/5)^35

So equalizing comparable terms on both sides we get m = 35
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Re: Exponents / powers [#permalink] New post 13 Mar 2014, 07:36
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chrish06 wrote:
How can I approach this question? Why is the answer 'D'?

If (1/5)m(1/4)18 = 1/2(10)35, m?
a. 18
b. 17
c. 21
d. 35
e. 3


Merging similar topics. Please refer to the solutions above and ask if anything remains unclear.

Theory on Exponents: math-number-theory-88376.html

All DS Exponents questions to practice: search.php?search_id=tag&tag_id=39
All PS Exponents questions to practice: search.php?search_id=tag&tag_id=60

Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html
Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html


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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ? [#permalink] New post 11 Jul 2014, 08:44
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ? [#permalink] New post 11 Jul 2014, 10:21
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I came across with this question lately during the test prep in the latest stage and I think is not a sub 600 level ??

someone can confirm me this impression ??

thanks


I think it's at most 620-630 level question.
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ? [#permalink] New post 11 Jul 2014, 11:44
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If (1/5)^m * (1/4)^18 = 1/(2(10)^35), then m = ? [#permalink] New post 23 Aug 2014, 22:26
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 :)
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Re: If (1/5)^m(1/4)^18=1/2(10)^35, then m = [#permalink] New post 31 Aug 2014, 04:33
GMATcrusher wrote:
Does anyone know how to solve this practice question from the free GMAT Pill practice set?

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

a) 17
b) 18
c) 34
d) 35
e) 36

Firstly simplify the equation on the left and bring it in the form of Prime Numbers as

(1/5)^m * (1/2)^36 as (1/4)^18= (1/2^2)^18
Equation on right-
1/2 *( 1/10)^35
= 1/2 * (1/2)^35 * (1/5)^35
=(1/2)^36 * (1/5)^35

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
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ? [#permalink] New post 07 Sep 2014, 23:41
\(\frac{1}{5^m} * \frac{1}{4^{18}} = \frac{1}{2*10^{35}}\)

Reciprocal the complete equation; Get rid of the fraction

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

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

m = 35

Answer = D
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m = 35 [#permalink] New post 02 Feb 2015, 13:31
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!!
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Re: m = 35 [#permalink] New post 02 Feb 2015, 14:16
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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!!

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 :-)
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Re: m = 35 [#permalink] New post 02 Feb 2015, 14:18
mikemcgarry 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!!

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!
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Re: m = 35 [#permalink] New post 02 Feb 2015, 14:23
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
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ? [#permalink] New post 04 Mar 2015, 05:17
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 \(1^m/5^m?\)
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ? [#permalink] New post 04 Mar 2015, 05:27
erikvm wrote:
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: 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..
Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?   [#permalink] 04 Mar 2015, 05:27

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(1/5)^m * (1/4)^18 = 1/2(10)^35 What is the value of m? sokenyou 0 11 Aug 2013, 00:32
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