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(1/5)^m (1/4)^10=1/2(10)^35, what is m?

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by 4GMAT_Mumbai » Thu Apr 15, 2010 10:12 pm
Hi Petenaz,

I am having trouble understanding the expression.

Could you please make it clearer with brackets and '*'s (to indicate multiplication).

Thanks.
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by petenaz » Fri Apr 16, 2010 8:04 am
this is the formula with the requested symbology: (1/5)^m*(1/4)^10=1/2(10)^35 so.... it is 1/5th to the power of m times 1/4th to the power of 10 equals 1 over(divided by) 2 times 10 to the power of 35. I hope this helps clear it up. Thank you for the reply!
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by akdayal » Sat Apr 17, 2010 1:47 am
m = -41
I am not getting any shot cut method at this time.
I am using conventional method to solve this problem
take log both side and solve for m

expression will become after simplifying

m = (21 log (2) - 35 )/log(5)

Hope this will help you.
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by nadaban » Mon Apr 19, 2010 11:28 am
petenaz wrote:this is the formula with the requested symbology: (1/5)^m*(1/4)^10=1/2(10)^35 so.... it is 1/5th to the power of m times 1/4th to the power of 10 equals 1 over(divided by) 2 times 10 to the power of 35. I hope this helps clear it up. Thank you for the reply!
Hi,

If you play around with the formula you get here (sort of finally :)):
5^(-m) = 2^54 * 5^35

From here you can see that there will not be a whole number solution, and you can go for the solution proposed by akdayal in the previous post and solve it with using the log.

Bests,
J
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by Brian@VeritasPrep » Mon Apr 19, 2010 1:53 pm
Hey guys,

I love this question...at least the version that I've seen on practice tests and elsewhere. I don't think that the question as given in this thread is a GMAT-style question, as no GMAT questions will require the explicit use of logarithms.

In the version I've seen, the key is to note that the term with the variable, (1/5)^m, is the only term that will supply the 35 required 5s to create the 10^35 that is in the denominator on the right hand side of the equation. Because that 10^35 also requires 35 2s to be (2*5)^35, and because there is an additional 2 in that denominator, the question that I've seen also has the (1/4) term taken to the 18th power, providing 36 2s.

So, for future reference in problems like these, you should first look to use prime factorization of exponential bases. Any question I've seen in this form has had that as its foundation.
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by petenaz » Tue Apr 20, 2010 9:44 am
actually, Brian is right. I was going from memory and mistyped the power of 18 on 1/4 as 10. So, I was wondering if Brian could show the steps to the final result. It would help to see each step defined if at all possible. The actual equation would be:

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

Sorry about the improper equation for those that were helping before.

Thanks for the help!
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by Brian@VeritasPrep » Tue Apr 20, 2010 1:16 pm
Glad to hear that, Petenaz - thanks for clarifying! Here's how I'd work that equation to solve for m:

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

First, I'd identify that the 4 on the left hand side and the 10 on the right hand side of the equation can be broken into their prime factors so that we have all prime bases for the exponents:

(1/5)^m * (1/2*2)^18 = 1/(2(2*5)^35)

Then, I'd multiply out all of the parentheses to simplify:

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

(note that all of the numerators of 1, when taken to any exponent, just stay as 1)

One more step in removing parentheses and streamlining the prime bases - that 2 in the denominator on the right is multiplied by 2^35, so that becomes 2^36:

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

Now, you can multiply out the 2^36 denominator from both sides, leaving:

1/5^m = 1/5^35

So m must be 35.

____________________________________________________________________

One kind of quick shortcut here is to, again using prime factors of exponent bases, recognize that the only way to get 10^35 on the right hand side of the equation is to have 35 2s and 35 5s on the left (10^35 = (2*5)^35). The only term that will give you any 5s is the 1/5^m term, so m has to be 35 for the equation to hold (assuming, as we did above, that the GMAT isn't using logarithms, etc. - and the answer choices, which are probably all integers, should confirm that assumption).
Brian Galvin
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by lightshare » Thu Dec 09, 2010 4:48 pm
Thank you! Thank you! Thank you! This one has been driving me nuts! I think this is too difficult to be the first question on the test!
Brian@VeritasPrep wrote:Glad to hear that, Petenaz - thanks for clarifying! Here's how I'd work that equation to solve for m:

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

First, I'd identify that the 4 on the left hand side and the 10 on the right hand side of the equation can be broken into their prime factors so that we have all prime bases for the exponents:

(1/5)^m * (1/2*2)^18 = 1/(2(2*5)^35)

Then, I'd multiply out all of the parentheses to simplify:

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

(note that all of the numerators of 1, when taken to any exponent, just stay as 1)

One more step in removing parentheses and streamlining the prime bases - that 2 in the denominator on the right is multiplied by 2^35, so that becomes 2^36:

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

Now, you can multiply out the 2^36 denominator from both sides, leaving:

1/5^m = 1/5^35

So m must be 35.

____________________________________________________________________

One kind of quick shortcut here is to, again using prime factors of exponent bases, recognize that the only way to get 10^35 on the right hand side of the equation is to have 35 2s and 35 5s on the left (10^35 = (2*5)^35). The only term that will give you any 5s is the 1/5^m term, so m has to be 35 for the equation to hold (assuming, as we did above, that the GMAT isn't using logarithms, etc. - and the answer choices, which are probably all integers, should confirm that assumption).
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