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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:

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Intern  Joined: 20 Feb 2012
Posts: 10
If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:  [#permalink]

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5 00:00

Difficulty:   5% (low)

Question Stats: 90% (01:18) correct 10% (01:52) wrong based on 230 sessions

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If $$(2^x) (2^y) = 8$$ and $$(9^x)(3^y) = 81$$, then (x,y) equals:

A. (1,2)
B. (2,1)
C. (1,1)
D. (2,2)
E. (1,3)

I understand how to cancel out things for the first half of the problem. For the second half, GMATPrep v2 says that (3^2x)(3^y) = 3^4 becomes 32x + y = 34. Can somebody explain how that happens?
Math Expert V
Joined: 02 Sep 2009
Posts: 64891
Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:  [#permalink]

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1
RyanP wrote:
If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:

A. (1,2)
B. (2,1)
C. (1,1)
D. (2,2)
E. (1,3)

I understand how to cancel out things for the first half of the problem. For the second half, GMATPrep v2 says that (3^2x)(3^y) = 3^4 becomes 32x + y = 34. Can somebody explain how that happens?

$$2^x*2^y=8$$ --> $$2^{x+y}=2^3$$ --> $$x+y=3$$;
$$9^x*3^y=81$$ --> $$3^{2x}*3^y=3^4$$ --> $$3^{2x+y}=3^4$$ --> $$2x+y=4$$;

Solving: $$x+y=3$$ and $$2x+y=4$$ we get $$x=1$$ and $$y=2$$.

Hope it's clear.
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Intern  Joined: 07 Jan 2011
Posts: 14
Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:  [#permalink]

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2
you could simply plug in answer choices into the equations and see which one is equal...
Intern  Joined: 24 Apr 2013
Posts: 40
Schools: Duke '16
Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:  [#permalink]

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If (2 to the power of X)(2 to the power of Y) and (9 to the power of X) (3 to the power of Y) = 81 then (X,Y) =

(1,2)

(2,1)

(1,1)

(2,2)

(1,3)
Math Expert V
Joined: 02 Sep 2009
Posts: 64891
Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:  [#permalink]

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If (2 to the power of X)(2 to the power of Y) and (9 to the power of X) (3 to the power of Y) = 81 then (X,Y) =

(1,2)

(2,1)

(1,1)

(2,2)

(1,3)

Merging similar topics. Please refer to the solutions above.
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Intern  Status: Currently Preparing the GMAT
Joined: 15 Feb 2013
Posts: 27
Location: United States
GMAT 1: 550 Q47 V23
GPA: 3.7
WE: Analyst (Consulting)
Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:  [#permalink]

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1
clicker wrote:
you could simply plug in answer choices into the equations and see which one is equal...

I think the plug-in method works best in the actual exam when you're fighting against the clock, or when you've exhausted all possible angles of attack on a notoriously difficult problem. But during preparation, I believe it is essential to understand how and why the concept work to solve these questions.

To solve this question you should be familiar with the following rules :

- $$a^x * a^y = a^(x+y)$$ (1) (notice that for this rule to work, we must have the same base, as in "a")
- $$(a^x)^y = a^(x*y)$$ (2)
- $$a^x = a^y$$ therefore $$x = y$$ (3) (same as with rule n°1, we need to have the same base)

Considering our problem, we have :

$$2^x*2^y = 8$$ which, according to (1) will yield $$2^(x+y) = 8 = 2^3$$ which, according to (3) will yield : $$x + y = 3$$
$$9^x*3^y = 81$$
Since $$9 = 3^2$$, then the above equation becomes : $$(3^2)^x*3^y = 81$$. According to (2), we can write : $$3^(2*x)*3^y = 81$$ which, according to (1) will yield $$3^(2x+y) = 81 = 3^4$$, which according to (3), will yield : $$2x + y = 4$$

We end up with a linear system of two equations :
$$x + y = 3$$
$$2x + y = 4$$

The solution of which is x = 1 and y = 2. Which is answer choice A.

Hope that helped Board of Directors D
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 5012
Location: India
GPA: 3.5
Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:  [#permalink]

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RyanP wrote:
If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:

A. (1,2)
B. (2,1)
C. (1,1)
D. (2,2)
E. (1,3)

I understand how to cancel out things for the first half of the problem. For the second half, GMATPrep v2 says that (3^2x)(3^y) = 3^4 becomes 32x + y = 34. Can somebody explain how that happens?

$$(2^x) (2^y) = 8$$

Or, $$2^{ x + y } = 2^3$$

So, $$x + y = 3$$-------------->(I)

$$(9^x)(3^y) = 81$$

Or, $$3^{2x + y } =3^4$$

So, $$2x + y = 4$$----------->(II)

Combine (I) and (II)

$$2x + 2y = 6$$
$$2x + y = 4$$

Solve for $$y$$

$$y = 2$$

Plug in y = 2 in ( I)

$$x + y = 3$$

Or, $$x + 2 = 3$$

So, $$x = 1$$

Hence (x , y ) = ( 1 , 2 )

Thus answer will be (A) (1,2)
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Director  G
Joined: 02 Sep 2016
Posts: 621
Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:  [#permalink]

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If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:

If base is same, we add powers.
So,
x+y= 3 (3 because 2^3=8 and we are comparing/equating powers)
2x+y= 4

Let's subtract the two equations:
2x+y=4
-x-y=-3

x= 1
and thus y=2

A. (1,2)
B. (2,1)
C. (1,1)
D. (2,2)
E. (1,3)
Intern  Joined: 06 Aug 2019
Posts: 3
Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:  [#permalink]

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Virgilius wrote:
clicker wrote:
you could simply plug in answer choices into the equations and see which one is equal...

I think the plug-in method works best in the actual exam when you're fighting against the clock, or when you've exhausted all possible angles of attack on a notoriously difficult problem. But during preparation, I believe it is essential to understand how and why the concept work to solve these questions.

To solve this question you should be familiar with the following rules :

- $$a^x * a^y = a^(x+y)$$ (1) (notice that for this rule to work, we must have the same base, as in "a")
- $$(a^x)^y = a^(x*y)$$ (2)
- $$a^x = a^y$$ therefore $$x = y$$ (3) (same as with rule n°1, we need to have the same base)

Considering our problem, we have :

$$2^x*2^y = 8$$ which, according to (1) will yield $$2^(x+y) = 8 = 2^3$$ which, according to (3) will yield : $$x + y = 3$$
$$9^x*3^y = 81$$
Since $$9 = 3^2$$, then the above equation becomes : $$(3^2)^x*3^y = 81$$. According to (2), we can write : $$3^(2*x)*3^y = 81$$ which, according to (1) will yield $$3^(2x+y) = 81 = 3^4$$, which according to (3), will yield : $$2x + y = 4$$

We end up with a linear system of two equations :
$$x + y = 3$$
$$2x + y = 4$$

(The solution of which is x = 1 and y = 2. Which is answer choice A.), Please solve how X=1 and Y=2

Hope that helped :-D Re: If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:   [#permalink] 07 Jun 2020, 01:27

# If (2^x) (2^y) = 8 and (9^x)(3^y) = 81, then (x,y) equals:   