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# PS - Exponents

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Manager
Joined: 12 Apr 2006
Posts: 210
Location: India

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24 May 2009, 05:45
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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

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Senior Manager
Joined: 08 Jan 2009
Posts: 317

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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.
Intern
Joined: 15 Feb 2009
Posts: 12

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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??
Intern
Joined: 17 May 2009
Posts: 10
Location: USA
Schools: Kellogs, Wharton, Chicago

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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.
Intern
Joined: 21 Apr 2009
Posts: 10
Location: DC

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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!
Intern
Joined: 15 Feb 2009
Posts: 12

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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

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: PS - Exponents   [#permalink] 30 May 2009, 14:07
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