It is currently 17 Oct 2017, 12:25

### 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

# PS - Exponents

Author Message
Manager
Joined: 12 Apr 2006
Posts: 213

Kudos [?]: 30 [0], given: 17

Location: India

### Show Tags

24 May 2009, 05:45
1
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x and y are integers and
[15^x + 15^(x+1)]/4^y = 15^y,

what is the value of x?
a) 2
b) 3
c) 4
d) 5
e) Cannot be determined

Kudos [?]: 30 [0], given: 17

Senior Manager
Joined: 08 Jan 2009
Posts: 326

Kudos [?]: 176 [0], given: 5

### Show Tags

24 May 2009, 06:32
Soln:

15 ^x + 15 ^(x+1) = 15^y * 4 ^y

15 ^x ( 16 ) = 60 ^ y

so 16 = 60^y/ 15 ^ x

so when y = 2 and x = 2 . 16 = 60^2 /15^2

so A.

Kudos [?]: 176 [0], given: 5

Intern
Joined: 15 Feb 2009
Posts: 12

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

### Show Tags

24 May 2009, 15:14
tkarthi4u wrote:
Soln:

15 ^x + 15 ^(x+1) = 15^y * 4 ^y

15 ^x ( 16 ) = 60 ^ y

so 16 = 60^y/ 15 ^ x

so when y = 2 and x = 2 . 16 = 60^2 /15^2

so A.

what happened with the exposents (y)?
dont we add exposents? so it gives 60^2y??

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

Intern
Joined: 17 May 2009
Posts: 10

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

Location: USA
Schools: Kellogs, Wharton, Chicago

### Show Tags

24 May 2009, 16:27
[15^x + 15^(x+1)]/(4^y) = 15^y
==> (15^x)*16 = (15^y)*(4^y)
==>(15^x)*(4^2) = (15^y)*(4^y)
Since 15 and 4 are relatively prime i.e 15 is not divisible by 4, the only way this equation is valid is when
y=2 and y=x, therefore x=2.

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

Intern
Joined: 21 Apr 2009
Posts: 10

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

Location: DC

### Show Tags

30 May 2009, 13:48
MontrealGirl wrote:
15^y * 4 ^y
= 60 ^ y

what happened with the exposents (y)?
dont we add exposents? so it gives 60^2y??

@MontrealGirl:
You only add the exponents when you are multiplying together items that have the exact same base. Thus, $$x^4 * x^5$$ is indeed $$x^9$$. This is one of the fundamental rules of exponents.

The rule that's used here, however, doesn't come up as often: When you multiply things together that happen to have the same exponent, then indeed multiply the old bases to get a new base, and then keep the exponent the same.

Look at, for example, $$2^5*3^5 = 6^5$$. You can even confirm that the left equals the right with a calculator.

Here's a way to think about it: The term $$15^y$$ means "$$y$$-many 15s, all multiplied together". Then the term $$4^y$$ means "$$y$$-many 4s, all multiplied together". Thus, what do you get when you stick those piles together? Imagine that each 15 "pairs up" with one of the 4s, so you have a bunch of 60s. How many? Exactly $$y$$ many, in fact.

An easy umbrella rule is that when you're multiplying (or dividing) and there are exponents involved, and something is the same, then whatever is the same will probably remain constant.

$$x^2 * x^3 = x^5$$
$$x^a * y^a = (xy)^a$$

Hope that explains it!

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

Intern
Joined: 15 Feb 2009
Posts: 12

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

### Show Tags

30 May 2009, 14:07
Liquidhypnotic wrote:
MontrealGirl wrote:
15^y * 4 ^y
= 60 ^ y

what happened with the exposents (y)?
dont we add exposents? so it gives 60^2y??

@MontrealGirl:
You only add the exponents when you are multiplying together items that have the exact same base. Thus, $$x^4 * x^5$$ is indeed $$x^9$$. This is one of the fundamental rules of exponents.

The rule that's used here, however, doesn't come up as often: When you multiply things together that happen to have the same exponent, then indeed multiply the old bases to get a new base, and then keep the exponent the same.

Look at, for example, $$2^5*3^5 = 6^5$$. You can even confirm that the left equals the right with a calculator.

Here's a way to think about it: The term $$15^y$$ means "$$y$$-many 15s, all multiplied together". Then the term $$4^y$$ means "$$y$$-many 4s, all multiplied together". Thus, what do you get when you stick those piles together? Imagine that each 15 "pairs up" with one of the 4s, so you have a bunch of 60s. How many? Exactly $$y$$ many, in fact.

An easy umbrella rule is that when you're multiplying (or dividing) and there are exponents involved, and something is the same, then whatever is the same will probably remain constant.

$$x^2 * x^3 = x^5$$
$$x^a * y^a = (xy)^a$$

Hope that explains it!

makes sense

thanks

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

Re: PS - Exponents   [#permalink] 30 May 2009, 14:07
Display posts from previous: Sort by