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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]

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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
[Reveal] Spoiler: OA

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

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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:

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

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

Since 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 = 1/(2*(10)^35), then m = ? [#permalink]

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New post 09 Feb 2012, 14:31
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 = 1/(2*(10)^35), then m = ? [#permalink]

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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 02 Apr 2013, 07:58
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: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ? [#permalink]

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New post 10 Aug 2013, 19:53
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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, 20:07, edited 1 time in total.

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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 Aug 2013, 01:32
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: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ? [#permalink]

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New post 13 Mar 2014, 08:36
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

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

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

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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 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 :)
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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 31 Aug 2014, 05:33
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]

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

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

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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!!


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]

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New post 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 \(1^m/5^m?\)

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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 04 Mar 2015, 06:27
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..
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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 04 Mar 2015, 06:33
Thanks for the quick response. I actually know these rules but I forget to apply them, its really frustrating :evil:

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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 04 Mar 2015, 06:36
erikvm wrote:
Thanks for the quick response. I actually know these rules but I forget to apply them, its really frustrating :evil:


doesnt matter erik... it happens with everyone if we don't refresh basics.... i am sure you will overcome all this by putting in some hardwork.. all the best..
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ? [#permalink]

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jst6059 wrote:
Question from the practice test that has me dumbfounded:

If \((\frac{1}{5})^m * (\frac{1}{4})^18 = (\frac{1}{2(10)^35})\), what does m equal?

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


Format your question properly and search for a question before you post. The question has already been discussed. Look above for the solution.

Topics merged.

As for your question, for all such questions, you need to break the bigger numbers down to their respective prime factors by noting that \(4=2^2\) and \(10=2*5\), After you are doing breaking both sides of the equation down to the respective prime factors, equate the powers of similar 'bases' to get to your final answer.

Also \(1/x = x^{-1}\) and \(x^a*x^b = x^{(a+b)}\)

Applying these relations to the question at hand you get:

\(5^{-m}*4^{-18} = 2^{-1}*10^{-35}\) ---> \(5^{-m}*(2^2)^{-18} = 2^{-1}*(2*5)^{-35}\) ---> \(5^{-m}*2^{-36} = 2^{-1}*(2^{-35}*5^{-35})\)

---> \(5^{-m}*2^{-36} = 2^{-1}*(2^{-35}*5^{-35})\) ---> \(5^{-m}*2^{-36} = 2^{-36}*5^{-35}\), equate both sides now to see that m=35. D is the correct answer.
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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 04 Oct 2015, 06:52
erikvm wrote:
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 \(1^m/5^m?\)


1) What is RHS / LHS?
LHS and RHS are the acronyms of Left hand side and RIgh hand side respectively.
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"?
We can go from 4^18 --> 2^36, because 4 is a power of two, (2^2 = 4)
Whereas 10 is not a power of 5. You cannot express 10 just by using 5.
Divide and multiply by 2 rules means to divide and multiply an expression by 2 to make it simpler.

For example: 18/5 = 18*2/5*2 = 36/10 = 3.6
By doing this, we can solve easily

3) 1/5^m - How do I get rid of the ^m in the numerator? I mean, isnt 1/5^m 1m/5m?..
Yes (1/5)^m = 1^m/5^m, but however many times you multiply 1, the result will always be 1.
Hence we can write 1^m simply as 1.
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Re: If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?   [#permalink] 04 Oct 2015, 06:52

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

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