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# If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha

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Joined: 09 Oct 2009
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If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha  [#permalink]

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Updated on: 06 Oct 2017, 11:11
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55% (hard)

Question Stats:

64% (01:49) correct 36% (02:10) wrong based on 889 sessions

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If x and y are integers and $$\frac{15^x + 15^{(x+1)}}{4^y} = 15^y$$ what is the value of x?

A. 2
B. 3
C. 4
D. 5
E. Cannot be determined

Okay so the correct answer is A, and below is the explanation given:
(15x + 15x+1) = 15y4y
[15x + 15x(151)] = 15y4y
(15x )(1 + 15) = 15y4y - HUH???
(15x)(16) = 15y4y
(3x)(5x)(24) = (3y)(5y)(22y)

Since both sides of the equation are broken down to the product of prime bases, the respective exponents of like bases must be equal.

2y = 4 so y = 2.
x = y so x = 2.

My question is how do we go from 15^x + (15^x)(15^1) to 15^x (1+15) Am I missing a major rule here? That step made no sense to me...

Originally posted by SnehaC on 18 Aug 2010, 07:33.
Last edited by Bunuel on 06 Oct 2017, 11:11, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha  [#permalink]

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18 Aug 2010, 07:48
23
6
SnehaC wrote:
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

Okay so the correct answer is A, and below is the explanation given:
(15x + 15x+1) = 15y4y
[15x + 15x(151)] = 15y4y
(15x )(1 + 15) = 15y4y - HUH???
(15x)(16) = 15y4y
(3x)(5x)(24) = (3y)(5y)(22y)

Since both sides of the equation are broken down to the product of prime bases, the respective exponents of like bases must be equal.

2y = 4 so y = 2.
x = y so x = 2.

My question is how do we go from 15^x + (15^x)(15^1) to 15^x (1+15) Am I missing a major rule here? That step made no sense to me...

I think the equation is $$\frac{(15^x + 15^{x+1})}{4^y} = 15^y$$

so now when you take $$15^x$$ common, then the equation becomes $$\frac{15^x( 1+ 15)}{4^y} = 15^y$$

$$15^x( 16) = 15^y4^y$$
$$15^x*4^2=15^y*4^y$$

Now, from the equation$$x=y=2$$

I hope that helps
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Re: If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha  [#permalink]

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18 Feb 2014, 20:05
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\frac{15^x( 1+ 15)}{4^y} = 15^y

(15^x . 4^2 ) / 4^y = 15^y. 4^0

15^x . 4^(2-y) = 15^y. 4^0

Equating the powers, x = y; 2-y = 0; So x = y = 2
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Re: If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha  [#permalink]

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06 Oct 2017, 10:52
SnehaC wrote:
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

We can simplify the given equation:

15^x + 15^x * 15 = 15^y * 4^y

15^x(1 + 15) = 15^y * 2^(2y)

15^x(2^4) = 15^y * 2^(2y)

Thus, we see that:

15^x = 15^y, so x = y

and

2^4 = 2^(2y), so 4 = 2y, or 2 = y.

Thus, x = 2.

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Re: If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha  [#permalink]

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15 Oct 2018, 15:55
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Re: If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha   [#permalink] 15 Oct 2018, 15:55
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# If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha

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