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# If 2^(x+y)*3^(x-y) = 32/3 then what is the value of product x and y?

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Joined: 31 Jan 2018
Posts: 9
If 2^(x+y)*3^(x-y) = 32/3 then what is the value of product x and y?  [#permalink]

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Updated on: 04 Feb 2018, 00:06
3
00:00

Difficulty:

15% (low)

Question Stats:

78% (01:28) correct 22% (02:22) wrong based on 78 sessions

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If $$2^{(x+y)} * 3^{(x-y)} = \frac{32}{3}$$ then what is the value of product x and y?

A. 2
B. 3
C. 5
D. 6
E. 16

Originally posted by ggpapas on 03 Feb 2018, 20:04.
Last edited by Bunuel on 04 Feb 2018, 00:06, edited 5 times in total.
formatted the question
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Re: If 2^(x+y)*3^(x-y) = 32/3 then what is the value of product x and y?  [#permalink]

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Updated on: 03 Feb 2018, 22:14
1
Thought so the question had a problem, hence removed the solution, sry

Since it is multiplication in the LHS, it becomes easy to equate LHS , RHS

2^(x+y) * 3^(x-y)= 2^5 * 3^-1

equating we get x+y=5 & x-y=-1
solving for x & y , we get 2&3 respt.
hence ans xy = 6
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Originally posted by doomedcat on 03 Feb 2018, 20:22.
Last edited by doomedcat on 03 Feb 2018, 22:14, edited 1 time in total.
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Re: If 2^(x+y)*3^(x-y) = 32/3 then what is the value of product x and y?  [#permalink]

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03 Feb 2018, 20:57
2
ggpapas wrote:
if 2^(x+y) + 3^(x-y) = 32/3 then what is the value of product x and y?

A.2
B.3
C.5
D.6
E.16

Hi ggpapas

The question seems to have an error in the highlighted part. It should be multiplication for the answer to be 6. Kindly confirm the source of the question

$$2^{(x+y)}*3^{(x-y)}=\frac{32}{3}=2^5*3^{-1}$$

$$=>x+y=5$$ & $$x-y=-1$$

therefore, $$x=2$$ & $$y=3$$

Hence $$x*y=2*3=6$$
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Re: If 2^(x+y)*3^(x-y) = 32/3 then what is the value of product x and y?  [#permalink]

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03 Feb 2018, 21:00
1
ggpapas wrote:
if 2^(x+y) + 3^(x-y) = 32/3 then what is the value of product x and y?

A.2
B.3
C.5
D.6
E.16

Hi ggpapas,

Can you please check if the question is correct. I tried for almost 45 minutes to figure out the answer but couldn't get any approach.
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Re: If 2^(x+y)*3^(x-y) = 32/3 then what is the value of product x and y?  [#permalink]

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03 Feb 2018, 21:24
1
ggpapas wrote:
if 2^(x+y) + 3^(x-y) = 32/3 then what is the value of product x and y?

A.2
B.3
C.5
D.6
E.16

It seems that there is a typo and it should be multiplication instead of addition. Please check once again.
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Re: If 2^(x+y)*3^(x-y) = 32/3 then what is the value of product x and y?  [#permalink]

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03 Feb 2018, 21:40
Hi all,

Thanks!
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Posts: 7978
Re: If 2^(x+y)*3^(x-y) = 32/3 then what is the value of product x and y?  [#permalink]

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03 Feb 2018, 21:52
ggpapas wrote:
if 2^(x+y) x 3^(x-y) = 32/3 then what is the value of product x and y?

A.2
B.3
C.5
D.6
E.16

$$2^{x+y}*3^{x-y}=\frac{32}{3}=\frac{2^5}{3}=2^5*3^{-1}$$..

So by equating powers
x+y=5 ...
x-y=-1.....x=y-1
So x+y=y-1+y=5.......2y=6....y=3
x=y-1=3-1=2..
xy=3*2=6

D
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Re: If 2^(x+y)*3^(x-y) = 32/3 then what is the value of product x and y?  [#permalink]

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03 Feb 2018, 23:38
ggpapas wrote:
if $$2^{(x+y)} * 3^{(x-y)} = \frac{32}{3}$$ then what is the value of product x and y?

A.2
B.3
C.5
D.6
E.16

Theory to solve such questions
The key to solving these kind of problems is to equate the left hand side to the
right hand side. Once you have a common base on both the side the exponents
can be equated.

For an equation $$Base1^{Exponent1} = Base2^{Exponent2}$$

If $$Base1 = Base2$$, then $$Exponent1 = Exponent 2$$

Now, coming back to this question.

$$2^{(x+y)} * 3^{(x-y)} = \frac{32}{3}$$ can be re-written as $$2^{(x+y)} * 3^{(x-y)} = 2^5 * 3^{-1}$$

Using this equation we get x+y = 5 and x-y = -1
Solving for x and y, we get x=2 and y=3
The product of x and y is 6(Option D)

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Re: If 2^(x+y)*3^(x-y) = 32/3 then what is the value of product x and y?  [#permalink]

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01 Aug 2019, 23:35
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Re: If 2^(x+y)*3^(x-y) = 32/3 then what is the value of product x and y?   [#permalink] 01 Aug 2019, 23:35
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