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Math Expert V
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For integers x and y,3^(4x+12)=5^(3x+y). What is the value of y?  [#permalink]

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Question Stats: 52% (01:36) correct 48% (01:57) wrong based on 241 sessions

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For integers x and y, 3^(4x+12)=5^(3x+y). What is the value of y?

A. -12
B. -3
C. 0
D. 9
E. Cannot be determined

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CEO  D
Status: GMATINSIGHT Tutor
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Re: For integers x and y,3^(4x+12)=5^(3x+y). What is the value of y?  [#permalink]

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Bunuel wrote:
For integers x and y, 3^(4x+12)=5^(3x+y). What is the value of y?

A. -12
B. -3
C. 0
D. 9
E. Cannot be determined

3^(4x+12)=5^(3x+y)

This relation will hold true if the power of 3 and 5 on both sides of equation become zero because $${anything}^0 = 1$$

i.e. 4x+12 = 0 and 3x+y=0
i.e. x = -2 and y = 9

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CEO  V
Joined: 12 Sep 2015
Posts: 3777
Re: For integers x and y,3^(4x+12)=5^(3x+y). What is the value of y?  [#permalink]

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Bunuel wrote:
For integers x and y, 3^(4x+12) = 5^(3x+y). What is the value of y?

A. -12
B. -3
C. 0
D. 9
E. Cannot be determined

The key word in this question is INTEGERS

Notice that, if x is an integer, then 4x+12 is an integer, which means 3^(4x+12) will equal the product of a bunch of 3's
Likewise, if x and y are integers, then 3x+y is an integer, which means 5^(3x+y) will equal the product of a bunch of 5's
Given these conditions, it seems impossible that 3^(4x+12) could ever equal 5^(3x+y)
HOWEVER, if the exponents 4x+12 and 3x + y both equal ZERO, then we get 3^0 and 5^0, and both of these evaluate to equal 1 - PERFECT!

So, let 4x+12 = 0 and let 3x+y = 0
Now we'll solve this system of equations for x and y.

First, if 4x+12 = 0, then x = -3
If x = -3, then we can take 3x+y = 0 and replace x with -3 to get: 3(-3) + y = 0
Simplify: -9 + y = 0
Solve: y = 9

So, x = -3 and y = 9, is a solution to the equation 3^(4x+12) = 5^(3x+y)

Cheers,
Brent
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Re: For integers x and y,3^(4x+12)=5^(3x+y). What is the value of y?  [#permalink]

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GMATinsight wrote:

i.e. 4x+12 = 0 and 3x+y=0
i.e. x = -2 and y = 9

Small error with the x-value above. Should be x = -3

Cheers,
Brent
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Re: For integers x and y,3^(4x+12)=5^(3x+y). What is the value of y?  [#permalink]

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1
Bunuel wrote:
For integers x and y, 3^(4x+12)=5^(3x+y). What is the value of y?

A. -12
B. -3
C. 0
D. 9
E. Cannot be determined

In order for the two sides of the equation to be equal, the exponent (4x+12) must equal zero and the exponent (3x+y) must equal zero, since 3^0 = 1 and 5^0 = 1. Thus:

4x + 12 = 0

4x = -12

x = -3

and

3x + y = 0

y = -3x

Since x = -3, y = -3(-3) = 9.

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Re: For integers x and y,3^(4x+12)=5^(3x+y). What is the value of y?  [#permalink]

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GMATinsight wrote:
Bunuel wrote:
For integers x and y, 3^(4x+12)=5^(3x+y). What is the value of y?

A. -12
B. -3
C. 0
D. 9
E. Cannot be determined

3^(4x+12)=5^(3x+y)

This relation will hold true if the power of 3 and 5 on both sides of equation become zero because $${anything}^0 = 1$$

i.e. 4x+12 = 0 and 3x+y=0
i.e. x = -2 and y = 9

Hello GMATinsight - You solution is really the quickest way to do! Somehow I spent 4 mins to solve this.

Just one correction, the value you mentioned for x is incorrect, x should be -3
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Retired Moderator P
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Re: For integers x and y,3^(4x+12)=5^(3x+y). What is the value of y?  [#permalink]

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For integers x and y, $$3^{4x+12}=5^{3x+y}$$. What is the value of y?

$$3^{4x+12}=5^{3x+y}$$

This is only possible when both 3 and 5 are raised to the power of ZERO.

4x + 12 = 0

3x + y = 0

Solving these two equations we get:

$$x = -3$$

$$y = 9$$

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"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."

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Re: For integers x and y,3^(4x+12)=5^(3x+y). What is the value of y?  [#permalink]

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Bunuel wrote:
For integers x and y, 3^(4x+12)=5^(3x+y). What is the value of y?

A. -12
B. -3
C. 0
D. 9
E. Cannot be determined

Since 3^(4x+12) = 5^(3x+y) and 3 and 5 are relatively prime to each other (that is, they don’t have a common factor other than 1), the only way those two expressions can be equal is if each base is raised to the zero power and thus each side is equal to 1.

Thus:

4x + 12 = 0

4x = -12

x = -3

Substituting, we have:

3(-3) + y = 0

-9 + y = 0

y = 9

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Posts: 1004
Location: India
Re: For integers x and y,3^(4x+12)=5^(3x+y). What is the value of y?  [#permalink]

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1
Bunuel wrote:
For integers x and y, 3^(4x+12)=5^(3x+y). What is the value of y?

A. -12
B. -3
C. 0
D. 9
E. Cannot be determined

Had thought of an approach, but still forgot to consider this

5^0 * 3^(4x+12)=5^(3x+y) 3^0

3x +y = 0 & 4x+12 = 0, x =-3

y = 3*3 = 9

D
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Manager  S
Joined: 22 Sep 2018
Posts: 249
Re: For integers x and y,3^(4x+12)=5^(3x+y). What is the value of y?  [#permalink]

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Bunuel wrote:
For integers x and y, 3^(4x+12)=5^(3x+y). What is the value of y?

A. -12
B. -3
C. 0
D. 9
E. Cannot be determined

My reasoning:

In order for these two values to be equal each other $$3^{(4x+12}) and 5^{(3x+y)}$$ must BOTH be equal to 1.

The reason is because 5 raised to any power will never equal 3 raised to any power unless they are both raised to the power of 0.

Explained even further: $$5^3$$ = 5*5*5. Notice how there are no 3s in the prime factorization.

So we can set 4x+12 = 0 and see that x = -3. If x =-3, y must equal 9
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Re: For integers x and y,3^(4x+12)=5^(3x+y). What is the value of y?  [#permalink]

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Bunuel wrote:
For integers x and y, 3^(4x+12)=5^(3x+y). What is the value of y?

A. -12
B. -3
C. 0
D. 9
E. Cannot be determined

In order for a power in base 3 to equal a power in base 5, the exponents of each base must equal zero. In other words, 3^0 = 1 and 5^0 = 1, and so 3^0 = 5^0. This is the only way that the original equation can be satisfied. Thus, since each exponent must equal 0, we have:
4x + 12 = 0

4x = -12

x = -3

Solving for y, we have:

3(-3) + y = 0

-9 + y = 0

y = 9

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If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: For integers x and y,3^(4x+12)=5^(3x+y). What is the value of y?   [#permalink] 06 Feb 2019, 20:51
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