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605-655 (Medium)|   Exponents|   Roots|      
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SnehaC
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If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha Updated on: 06 Oct 2017, 11:11
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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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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45% (medium)

Question Stats:

69% (01:46) correct 31% (02:14) wrong based on 2092 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

Show Spoiler
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...
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A
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cipher
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If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha 18 Aug 2010, 07:48
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SnehaC
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

Show Spoiler
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...
Show more

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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PareshGmat
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If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha 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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If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha 06 Oct 2017, 10:52
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SnehaC
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
Show more

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.

Answer: A
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If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha 15 Apr 2021, 07:02
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SnehaC
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

Show more



hello BrentGMATPrepNow :) ok this how i tried solved it and got stuck ...i know it is boring story ... :lol:

i would appreciate if you could explain what am i doing wrong and why :grin: :)

\(\frac{15^x + 15^{(x+1)}}{4^y} = 15^y\)

\(\frac{15^x + 15^x*15^1}{4^y} = 15^y\) here i factored out \(15^x\) from \(( 15^x+15^x) \)

\(\frac{15^x (1+1) *15^1}{4^y} = 15^y\)

\(15^x (2) *15^1 = 15^y 4^y\)

\(15^x (2) *15^1 = 60^y\)
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If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha 15 Apr 2021, 07:41
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dave13
SnehaC
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




hello BrentGMATPrepNow :) ok this how i tried solved it and got stuck ...i know it is boring story ... :lol:

i would appreciate if you could explain what am i doing wrong and why :grin: :)

\(\frac{15^x + 15^{(x+1)}}{4^y} = 15^y\)

\(\frac{15^x + 15^x*15^1}{4^y} = 15^y\) here i factored out \(15^x\) from \(( 15^x+15^x) \)

\(\frac{15^x (1+1) *15^1}{4^y} = 15^y\)

\(15^x (2) *15^1 = 15^y 4^y\)

\(15^x (2) *15^1 = 60^y\)
Show more

I have highlighted the error above.
Here's an analogous mistake: 3 + 3 * 2 = 3(1 + 1) * 2
As you can see, the two calculations are not equivalent.
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If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha 15 Apr 2021, 07:53
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dave13
SnehaC
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




hello BrentGMATPrepNow :) ok this how i tried solved it and got stuck ...i know it is boring story ... :lol:

i would appreciate if you could explain what am i doing wrong and why :grin: :)

\(\frac{15^x + 15^{(x+1)}}{4^y} = 15^y\)

\(\frac{15^x + 15^x*15^1}{4^y} = 15^y\) here i factored out \(15^x\) from \(( 15^x+15^x) \)

\(\frac{15^x (1+1) *15^1}{4^y} = 15^y\)

\(15^x (2) *15^1 = 15^y 4^y\)

\(15^x (2) *15^1 = 60^y\)

I have highlighted the error above.
Here's an analogous mistake: 3 + 3 * 2 = 3(1 + 1) * 2
As you can see, the two calculations are not equivalent.
Show more


BrentGMATPrepNow thanks! any tips how i should avoid such mistakes and when i can safely factor out ? :)
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If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha 15 Apr 2021, 09:34
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BrentGMATPrepNow thanks! any tips how i should avoid such mistakes and when i can safely factor out ? :)
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I think the best way to avoid making any mistakes is to fully understand why the original mistake was incorrect in the first place.
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If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha 29 Apr 2021, 19:00
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If x and y are integers and (15^x + 15^(x+1))/4^y = 15^y wha 02 Dec 2025, 02:04
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