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Which of the following best approximates the value of q if

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VP
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Which of the following best approximates the value of q if [#permalink] New post 08 Aug 2006, 17:43
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A
B
C
D
E

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Which of the following best approximates the value of q if 5^28+3^11=5^q ?

A 39
B 30
C 28
D 27
E 17


please explain
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 [#permalink] New post 08 Aug 2006, 18:45
Please explain ASAP Haha
Really, if I could understand how to factor these questions, I would have a 50 math score. I still have yet to find a good source on the net on how to do these. I had limited math as undergrad, only bus calc, so some of this is puzzling...
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 [#permalink] New post 08 Aug 2006, 20:49
ok.. After trying for more than 20 minutes... I have no explanation but the following.

Answer: B : 30

q has to be greater than 28 otherwise 3^11 ~ approximates to zero.

Eliminating A. 39 because.. no idea.

Sorry thats the best I could not. I am not that smart... :(
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 [#permalink] New post 08 Aug 2006, 21:00
haas_mba07 wrote:
q has to be greater than 28 otherwise 3^11 ~ approximates to zero.(


are you sure? seems you have not taken fruit.. :wink:

ok. q would be 29, if the equation is as under: 5^28 + [5 (5^28)] = 5^q?
q would be 30, if the equation is as under: 5^28 + [24 (5^28)] = 5^q?

note the red part.
in first eq... how bigger is 5 (5^28) than 3^11?

hope its clear.
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 [#permalink] New post 08 Aug 2006, 22:42
C

3^11 is too small as compared to 5^28

3^11 have 8 or 9 digits
5^28 have around 25 digits.
So 3^11 is billionth billionth of 5^28.

Hence q will be 28.
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 [#permalink] New post 09 Aug 2006, 02:41
here is a more detailed explanation,

3^11 = 3^3*3^3*3^3*3^2..

and .3^3 = 27 approx 25 ie: 5^2

therfore 3^11 approx =( 5^2)^3 * 3^2.... substitute

5^28+3^11 = 5^28+5^6*3^2 ie: 5^6(5^22 + 3^2) if we compared 3^2 (ie: 9) to such a huge exponent as 5^22 we can see that it approximately has negligable effect if added to it

Thus 5^22+3^2 approx = 5^22

So, 5^6(5^22 + negligabel) = 5^28

therfore 5^28+3^11 = 5^28

Hence q is approx = 28

I hope i helped.....take care
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 [#permalink] New post 09 Aug 2006, 04:09
Hi Haas , i think you ve got a very valid point of view about approximation,

However when we approximate we should always ask ourslves we are approximating .............compared to what?????

ie: if we are approximating 0.42

We actually compared 2 to ten so we answere 0.42 is approx = 0.4

Similarly if we compared 3^11 to 5^28 we have a bigger (exponantial)reason to neglect the 3^11.

Take care.
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 [#permalink] New post 09 Aug 2006, 08:48
I don't think any amount of fruit would have helped me on that one... :P

Professor wrote:
are you sure? seems you have not taken fruit.. :wink:
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 [#permalink] New post 09 Aug 2006, 13:05
:lol: Hass ...trust me when i tell you that a day after day i face more than 10 to 15 Q problems that i dont have a reply to except a very big big smile

We are learning and trying to beat the Gmat all mighty. and sure we will.

Take care
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 [#permalink] New post 10 Aug 2006, 01:36
5^28 + 3^11 = 5^q
so 28 log5 + 11 log3 = qlog5
(q-28) log5 = 11log3

log5 is 1.6 (log5 to the base e) log3 is approx 1.1
log5 is .7(log5 to the base of 10) log3 is .5(log3 to the base of 10)




solving this i am getting q is approx 35 or 36.
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 [#permalink] New post 10 Aug 2006, 01:44
C 28.

3^11 is too small compared to 5^28. hence adding 5^28 + 3^11 wont be much greater than 5^28.
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 [#permalink] New post 10 Aug 2006, 06:25
5^28 + 3^11 = 5^q ( therefore q is >0 and greater than 28)

3^11 = 5^q - 5^28

3^11 = 5^28*(5^(q-28)-1) (since q >28 from above)

1+3^11/5^28=5^(q-28)

3^11/5^28 = approx(0) ( since 5^28 is so much larger than 3^11)

1 = 5^(q-28)

q - 28 = approx (0)

q = 28.
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 [#permalink] New post 10 Aug 2006, 08:06
god..i cant belive some of u r doing logrithmic calculations on this problem..this question shud be done under 12 sec....

5^28 is considerably bigger than 3^11...therefore q=28

30 will be too large...

C it is..
  [#permalink] 10 Aug 2006, 08:06
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