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PS - Exponents

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PS - Exponents [#permalink]

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New post 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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Re: PS - Exponents [#permalink]

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New post 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.

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Re: PS - Exponents [#permalink]

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New post 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??

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Re: PS - Exponents [#permalink]

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New post 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.

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Re: PS - Exponents [#permalink]

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New post 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!

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Re: PS - Exponents [#permalink]

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New post 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

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