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y in terms of x

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by aatech » Thu May 15, 2008 7:15 am
The question does not look correct... Can you please verify
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by gibran » Thu May 15, 2008 7:27 am
Sorry, here is the question again .....

If 5^x – 5^x - 3 = (124)(5^y), what is y in terms of x?
A. x
B. x - 6
C. x - 3
D. 2x + 3
E. 2x + 6
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by punit.kaur.mba » Thu May 15, 2008 7:33 am
Still looks incorrect..
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by aatech » Thu May 15, 2008 7:34 am
Just to clarify for others Q should be

5^x - 5^[x-3] = 124 * 5^y

=> 5^x - [5^x/5^3] = 124*5^y

=> 5^x[1 - 1/125] = 124*5^y

=> [5^x * 124]/125 = 124*5^y

=> 5^[x-3]*124 = 124*5^y

so, y = x-3

ANS C
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by gibran » Thu May 15, 2008 7:42 am
Thanks a lot :) Very clever thinking!!!!!
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by neilcao » Wed Dec 08, 2010 8:06 am
Sorry how did you get this ,can u break it a bit further please?

=> 5^x[1 - 1/125] = 124*5^y

=> [5^x * 124]/125 = 124*5^y

5^x[1 - 1/125] became [5^x * 124]/125

Thanks
aatech wrote:Just to clarify for others Q should be

5^x - 5^[x-3] = 124 * 5^y

=> 5^x - [5^x/5^3] = 124*5^y

=> 5^x[1 - 1/125] = 124*5^y

=> [5^x * 124]/125 = 124*5^y

=> 5^[x-3]*124 = 124*5^y

so, y = x-3

ANS C
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by neilcao » Wed Dec 08, 2010 8:09 am
Sorry worked it out, saw it wrong.

neilcao wrote:Sorry how did you get this ,can u break it a bit further please?

=> 5^x[1 - 1/125] = 124*5^y

=> [5^x * 124]/125 = 124*5^y

5^x[1 - 1/125] became [5^x * 124]/125

Thanks
aatech wrote:Just to clarify for others Q should be

5^x - 5^[x-3] = 124 * 5^y

=> 5^x - [5^x/5^3] = 124*5^y

=> 5^x[1 - 1/125] = 124*5^y

=> [5^x * 124]/125 = 124*5^y

=> 5^[x-3]*124 = 124*5^y

so, y = x-3

ANS C
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