Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 05 May 2015, 11:56

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

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
00:00

Difficulty:

(N/A)

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

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.

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

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
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!

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!

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!

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
Similar topics Replies Last post
Similar
Topics:
2 What does k equal if (3k)^x=1 1 04 Aug 2013, 02:36
What is the value of v^3 k^3? (1) vk > 0 (2) v - k = 6 0 18 Dec 2010, 13:26
4 What is the value of v^3 - k^3 ? 5 04 Aug 2009, 05:13
3 If 3^k + 3^k = (3^9)^(3^9) - 3^k, then k = ? 11 17 Jan 2006, 11:06
What is the value of v(exp)3 k(exp) 3? (1) v k > 0 (2) v 1 10 Oct 2005, 05:23
Display posts from previous: Sort by