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If 3^k+3^k=((3^9)^3)^9 -3^k, what is the

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If 3^k+3^k=((3^9)^3)^9 -3^k, what is the [#permalink] New post 07 Sep 2006, 05:51
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If 3^k+3^k=((3^9)^3)^9 -3^k, what is the value of k?

a. 243
b.3^{11}-1
c. 3^{10}-1
d. 3^{12}-1
e. 3^{10}-2
f. none of the above

Answer is not F.

Last edited by GMATT73 on 08 Sep 2006, 01:39, edited 1 time in total.
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 [#permalink] New post 08 Sep 2006, 04:21
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Thanks for posting the clarification GMATT73...

I corrected my answer to :

b. (3^11)-1

3^(k+1) = 3^(3^2 x 3^9)
or k+1 = 3^11
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 [#permalink] New post 07 Sep 2006, 06:13
A

3^K = ((3^9)^3)^9

9*3*9 = 243

3^k = 3^243

k = 243

A
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 [#permalink] New post 07 Sep 2006, 08:02
F

3^k+3^k={[(3^9)]^3}^9 -3^k
can be written as

3^k+3^k+3^k={[(3^9)]^3}^9

3^(k+1) = 3^243
k+1 = 243
k = 242 or (3^5)-1
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 [#permalink] New post 07 Sep 2006, 08:04
f. none of the above

Simplifying the given eqn:

2.3^k = 3^243 - 3^k
3^(k+1) = 3^243

Or k+1 = 243
k=242.

Answer: F
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 [#permalink] New post 07 Sep 2006, 10:48
3^k+3^k={[(3^9)]^3}^9 -3^k

3^k *2 = 3^243 - 3^k

3^k * 2 =3^k(3^(243-k) - 1)

thus 3^(243-k) - 1 = 2 thus 3^(243-k) = 3^1

therfore 243-k =1

ie k=242

thus answer should be F
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Re: PS Raising it to the next level [#permalink] New post 07 Sep 2006, 10:50
No.. its not ...

Its of the form

[(a^b)^c]^d - e^f = a^(bcd) - e^f

3^(243 - 3^k) = 3^243/3^(3k) in which case you can cross multiply and get your equation.

But it is actually,

3^243 - 3^k in which case you would move it to the LHS and get 3^k+3^k+3^k = 3 x 3^k = 3^(k+1)



gk3.14 wrote:

shouldn't this be 2.3^k.3^k? It is 3^(243 - 3^k)


GMATT73 wrote:
If 3^k+3^k={[(3^9)]^3}^9 -3^k,
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 [#permalink] New post 07 Sep 2006, 10:54
I think the bold part is wrong it should be

3^k[3^k + 3^k]=2(3^k^2)


ps_dahiya wrote:
F

3^k+3^k={[(3^9)]^3}^9 -3^k
can be written as

3^k+3^k+3^k={[(3^9)]^3}^9

3^(k+1) = 3^243
k+1 = 243
k = 242 or (3^5)-1
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 [#permalink] New post 07 Sep 2006, 10:58
3^k + 3^k= 3^(243 - 3^3+k)
= 3^243/ 3^(3^3+k)
= 3^243/3^(9+3k)
= 3^234/ 3^3k
(3^k + 3^k) 3^(3K) = 3^ 234
2.3^4k = 3^234

This is as far as I got.. can someone tell me what I am doing different from you all..

Thanks!
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Re: PS Raising it to the next level [#permalink] New post 07 Sep 2006, 11:00
agreed...but i thnk the question says 3^[9-3k]

haas_mba07 wrote:
No.. its not ...

Its of the form

[(a^b)^c]^d - e^f = a^(bcd) - e^f

3^(243 - 3^k) = 3^243/3^(3k) in which case you can cross multiply and get your equation.

But it is actually,

3^243 - 3^k in which case you would move it to the LHS and get 3^k+3^k+3^k = 3 x 3^k = 3^(k+1)



gk3.14 wrote:

shouldn't this be 2.3^k.3^k? It is 3^(243 - 3^k)


GMATT73 wrote:
If 3^k+3^k={[(3^9)]^3}^9 -3^k,
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Re: PS Raising it to the next level [#permalink] New post 07 Sep 2006, 12:08
I don't think it does, unless it was mis-typed...

3^k+3^k={[(3^9)]^3}^9 -3^k ........... (1)

is definately different from

3^k+3^k={[(3^9)]^3}^(9 -3^k).............. (2)

Those would be two different equations.

Eqn (1) would be F.

However, eqn 2 when solved gives:

2 x 3^k = ((3^27)^(9-3^k)

2 x 3^k = (3^243)/(3^81k)

2 x 3^82k = 3^243

This has no solution... F again.


So unless there is a pair of ( )'s to make it clearer, it would be a guess to think either way.



fresinha12 wrote:
agreed...but i thnk the question says 3^[9-3k]

haas_mba07 wrote:
No.. its not ...

Its of the form

[(a^b)^c]^d - e^f = a^(bcd) - e^f

3^(243 - 3^k) = 3^243/3^(3k) in which case you can cross multiply and get your equation.

But it is actually,

3^243 - 3^k in which case you would move it to the LHS and get 3^k+3^k+3^k = 3 x 3^k = 3^(k+1)



gk3.14 wrote:

shouldn't this be 2.3^k.3^k? It is 3^(243 - 3^k)


GMATT73 wrote:
If 3^k+3^k={[(3^9)]^3}^9 -3^k,
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 [#permalink] New post 08 Sep 2006, 01:40
Putting this back at the top of the board. Check the attachment for clarification. Answer is not F.
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 [#permalink] New post 08 Sep 2006, 04:25
haas_mba07 wrote:
Thanks for posting the clarification GMATT73...

I corrected my answer to :

b. (3^11)-1

3^(k+1) = 3^(3^2 x 3^9)
or k+1 = 3^11


where did I cause all the confusion in the original?
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 [#permalink] New post 08 Sep 2006, 04:51
Nope.. you didn't!
My bad... :-)


GMATT73 wrote:
haas_mba07 wrote:
Thanks for posting the clarification GMATT73...

I corrected my answer to :

b. (3^11)-1

3^(k+1) = 3^(3^2 x 3^9)
or k+1 = 3^11


where did I cause all the confusion in the original?
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 [#permalink] New post 08 Sep 2006, 05:16
haas_mba07 wrote:
Nope.. you didn't!
My bad... :-)


GMATT73 wrote:
haas_mba07 wrote:
Thanks for posting the clarification GMATT73...

I corrected my answer to :

b. (3^11)-1

3^(k+1) = 3^(3^2 x 3^9)
or k+1 = 3^11


where did I cause all the confusion in the original?


Pretty straightforward when you see the problem in exponential form, eh?
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 [#permalink] New post 08 Sep 2006, 05:17
Absolutely.. I got messed up in reading the question all the brackets.

Need to pay more attention..

BTW, I posted a set of SCs, might interest you.

GMATT73 wrote:
haas_mba07 wrote:
Nope.. you didn't!
My bad... :-)


GMATT73 wrote:
haas_mba07 wrote:
Thanks for posting the clarification GMATT73...

I corrected my answer to :

b. (3^11)-1

3^(k+1) = 3^(3^2 x 3^9)
or k+1 = 3^11


where did I cause all the confusion in the original?


Pretty straightforward when you see the problem in exponential form, eh?
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 [#permalink] New post 24 Sep 2006, 03:09
Uhm, I think it's B
I used logarithm, thus: k+1=(3^9) * 9
=> B
  [#permalink] 24 Sep 2006, 03:09
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