Author
Message
TAGS:
Current Student

Joined: 29 Jan 2005

Posts: 5244

Followers: 23

Kudos [? ]:
179
[0 ] , given: 0

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

Question Stats:

0% (00:00) correct

0% (00:00) wrong

based on 0 sessions
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.

Intern

Joined: 04 May 2006

Posts: 40

Followers: 0

Kudos [? ]:
0
[0 ] , given: 0

A
3^K = ((3^9)^3)^9
9*3*9 = 243
3^k = 3^243
k = 243
A

CEO

Joined: 20 Nov 2005

Posts: 2913

Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

Followers: 18

Kudos [? ]:
121
[0 ] , given: 0

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

_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

VP

Joined: 02 Jun 2006

Posts: 1267

Followers: 2

Kudos [? ]:
42
[0 ] , given: 0

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

SVP

Joined: 05 Jul 2006

Posts: 1519

Followers: 5

Kudos [? ]:
115
[0 ] , given: 39

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

VP

Joined: 02 Jun 2006

Posts: 1267

Followers: 2

Kudos [? ]:
42
[0 ] , given: 0

Re: PS Raising it to the next level [#permalink ]
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 ,

Current Student

Joined: 28 Dec 2004

Posts: 3391

Location: New York City

Schools: Wharton'11 HBS'12

Followers: 13

Kudos [? ]:
181
[0 ] , given: 2

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

Senior Manager

Joined: 14 Aug 2006

Posts: 365

Followers: 1

Kudos [? ]:
3
[0 ] , given: 0

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!

Current Student

Joined: 28 Dec 2004

Posts: 3391

Location: New York City

Schools: Wharton'11 HBS'12

Followers: 13

Kudos [? ]:
181
[0 ] , given: 2

Re: PS Raising it to the next level [#permalink ]
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 ,

VP

Joined: 02 Jun 2006

Posts: 1267

Followers: 2

Kudos [? ]:
42
[0 ] , given: 0

Re: PS Raising it to the next level [#permalink ]
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 ,

Current Student

Joined: 29 Jan 2005

Posts: 5244

Followers: 23

Kudos [? ]:
179
[0 ] , given: 0

Putting this back at the top of the board. Check the attachment for clarification. Answer is not F.

VP

Joined: 02 Jun 2006

Posts: 1267

Followers: 2

Kudos [? ]:
42
[1 ]
, given: 0

1

This post received KUDOS

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

Current Student

Joined: 29 Jan 2005

Posts: 5244

Followers: 23

Kudos [? ]:
179
[0 ] , given: 0

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?

VP

Joined: 02 Jun 2006

Posts: 1267

Followers: 2

Kudos [? ]:
42
[0 ] , given: 0

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?

Current Student

Joined: 29 Jan 2005

Posts: 5244

Followers: 23

Kudos [? ]:
179
[0 ] , given: 0

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?

VP

Joined: 02 Jun 2006

Posts: 1267

Followers: 2

Kudos [? ]:
42
[0 ] , given: 0

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?

Manager

Joined: 06 Feb 2006

Posts: 223

Followers: 2

Kudos [? ]:
4
[0 ] , given: 0

Uhm, I think it's B
I used logarithm, thus: k+1=(3^9) * 9
=> B