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tough one!

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Source: — Data Sufficiency |

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by Anurag@Gurome » Wed Feb 09, 2011 10:08 pm
3^7 - 3^5 = (2^x)(3^y) implies 3^5 (3^2 - 1) = (2^x)(3^y)
3^5 (9 - 1) = (2^x)(3^y)
3^5(8) = (2^x)(3^y)
(2^3)(3^5) = (2^x)(3^y)
On comparing both the sides (since the bases are same so the powers will also be the same),
x = 3 and y = 5
Hence, x + y = 3 + 5 = [spoiler]8 [/spoiler]

BTW, the question is posted under incorrect forum, should have been posted under Problem Solving.
Last edited by Anurag@Gurome on Wed Feb 09, 2011 10:10 pm, edited 1 time in total.
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by sallywoo » Wed Feb 09, 2011 10:10 pm
wow! thank you!
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by Anurag@Gurome » Wed Feb 09, 2011 10:11 pm
sallywoo wrote:wow! thank you!
You are welcome!
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by Night reader » Thu Feb 10, 2011 1:12 am
sallywoo wrote:my tutor has been e-mailing me tough quant questions, and he just sent me this one. any ideas?

If 3^7 - 3^5 = (2^x)(3^y), what is the value of x + y?

thanks!
30 sec. ---> 3^5(3^2-1)=(2^x)(3^y), (3^5)(2^3)=(2^x)(3^y), (3^y)/3^5=(2^3)/(2^x), 3^(y-5)=2^(3-x), y-5(>) (<) (=) 3-x, x+y(>)(<)(=)8

actually unsolvable, missing data

@Anurag, I am hesitating to disagree with your solution (I looked after solved :( )
x and y can be real numbers, i.e. (ir)rational too; why we make assumption of the powers being integer type. Instead I want to see the answer choices to select A-E.
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by Anurag@Gurome » Thu Feb 10, 2011 5:24 am
Night reader wrote:actually unsolvable, missing data

@Anurag, I am hesitating to disagree with your solution (I looked after solved :( )
x and y can be real numbers, i.e. (ir)rational too; why we make assumption of the powers being integer type.
Yes, you're correct.
Unless x and y are integers, there will be infinite possible values for them. I shouldn't have assumed them as integers without the options.
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