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# If x and y are positive integers and 5^x

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If x and y are positive integers and 5^x  [#permalink]

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06 Apr 2012, 09:42
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If x and y are positive integers and $$(5^x)-(5^y)=(2^{y-1})*(5^{x-1})$$, what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12
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If x and y are positive integers and 5^x  [#permalink]

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06 Apr 2012, 14:01
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BN1989 wrote:
If x and y are positive integers and (5^x)-(5^y)=(2^(y-1))*(5^(x-1)), what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

Notice that we are told that $$x$$ and $$y$$ are positive integers.

$$5^x-5^y=2^{y-1}*5^{x-1}$$;

$$5^x-2^{y-1}*5^{x-1}=5^y$$;

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

$$5^x(10-2^y)=2*5^{y+1}$$.

Now, since the right hand side is always positive then the left hand side must also be positive, hence $$10-2^y$$ must be positive, which means that $$y$$ can take only 3 values: 1, 2 and 3.

By trial and error we can find that only $$y=3$$ gives integer value for $$x$$:

$$5^x(10-2^3)=2*5^{3+1}$$;

$$2*5^x=2*5^4$$;

$$x=4$$ --> $$xy=12$$.

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Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x  [#permalink]

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19 Feb 2014, 22:50
8
5
Faster: First constrain the possible answers. We know $$x>y$$ since if $$x=y$$ then the left-hand side is 0 or if $$x<y$$ then the LHS is negative... but the RHS is always positive. Now act: divide both sides by $$5^{x-1}$$ to get $$5-5^{y-x+1} = 2^{y-1}$$.

Since the new RHS is a power of two, the LHS must equal 1, 2, or 4. The only power of 5 that gets us one of those is $$5-5^0=5-1=4$$. That means $$y=3$$ and thus $$x=4$$.

Oh, and BTW: The thread title is different than the equality in your post. (Title should have 5^{x-1}, not 5^x.)
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Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x  [#permalink]

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19 Feb 2014, 22:25
21
7
MrWallSt wrote:
If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^(x-1), what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

A little bit of observation can help you solve this question within a minute.

x and y are positive integers which means we will have clean numbers. On the right hand side, you have a 2 as a factor while it is not there on the left hand side. Can a 2 be generated on the left hand side by the subtraction? Here I am thinking that if we take 5^y common on the left hand side, I might be able to get a 2.

$$5^y (5^{x-y} - 1) = 2^{y-1}*5^{x-1}$$
Now I want only 2s and 5s on the left hand side. If x-y is 1, then $$(5^{x-y} - 1)$$ becomes 4 which is 2^2. If instead x - y is 2 or more, I will get factors such as 3, 13 too. So let me try putting x - y = 1 to get
$$5^y (2^2) = 2^{y-1}*5^{x-1}$$

This gives me y - 1 = 2
y = 3
x = 4
Check to see that the equations is satisfied with these values. Hence xy = 12

Note that it is obvious that y is less than x and that is the reason we took $$5^y$$ common. The reason it is obvious is that the right hand side is positive. So the left hand side must be positive too. This means $$5^x > 5^y$$ which means x > y.
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Re: If x and y are positive integers and 5^x  [#permalink]

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24 Sep 2013, 23:04
2
danzig wrote:
Bunuel wrote:
BN1989 wrote:
If x and y are positive integers and (5^x)-(5^y)=(2^(y-1))*(5^(x-1)), what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

Notice that we are told that $$x$$ and $$y$$ are positive integers.

$$5^x-5^y=2^{y-1}*5^{x-1}$$ --> $$5^x-2^{y-1}*5^{x-1}=5^y$$ --> $$5^x(1-\frac{2^y}{2}*\frac{1}{5})=5^y$$ --> $$5^x(10-2^y)=2*5^{y+1}$$.

Now, since the right hand side is always positive then the left hand side must also be positive, hence $$10-2^y$$ must be positive, which means that $$y$$ can take only 3 values: 1, 2 and 3.

By trial and error we can find that only $$y=3$$ gives integer value for $$x$$: $$5^x(10-2^3)=2*5^{3+1}$$ --> $$2*5^x=2*5^4$$ --> $$x=4$$ --> $$xy=12$$.

Bunuel, I have a question, how did you know that you had to reorganize the equation in this way?

From $$5^x-5^y=2^{y-1}*5^{x-1}$$ TO $$5^x-2^{y-1}*5^{x-1}=5^y$$

Thanks!

Not directed at me, However you can re-arrange it another way.

We have $$5^x-5^y = 2^{y-1}*5^{x-1}$$, Dividing on both sides by $$5^{x-1}$$, we have
$$5-5^{y+1-x} = 2^{y-1} \to 5 = 5^{y+1-x} + 2^{y-1}$$. Now, as x and y are positive integers, the only value which $$2^{y-1}$$ can take is 4.Thus, y-1 = 2, y = 3. Again, the value of $$5^{y+1-x}$$ has to be 1, thus, y+1-x = 0 $$\to$$ x = y+1 = 3+1 = 4. Thus, x*y = 4*3 = 12.

Thus, I think you could re-arrange in any order, as long as you get a tangible logic.
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Re: If x and y are positive integers and 5^x  [#permalink]

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20 Feb 2014, 04:53
2
BN1989 wrote:
If x and y are positive integers and $$(5^x)-(5^y)=(2^{y-1})*(5^{x-1})$$, what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

by dividing both sides of the equation with 5^x we have 1- 5^(y-x) = 2^(y-1) * 5^(x-1-x)

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

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

now minimum value of 2^(y-1) =1, hence 2^(y-1) must be equal to 4 and 5^(y-x+1) must be equal to 1 for the R.H.S to become equal to L.H.S.

2^(y-1) = 4 for y=3
and 5^(y-x+1) =1 for x=4 (as y=3)

hence product of xy = 12
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Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x  [#permalink]

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19 Feb 2014, 23:01
1
@Karishma, thanks again.

@SizeTrader, appreciate the solution and that reasoning was excellent. Also, the reason the title says 5^x is because I reached the character limit for the title and it got cut off
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Re: If x and y are positive integers and 5^x  [#permalink]

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20 Feb 2014, 15:02
1
If x and y are positive integers and (5^x)-(5^y)=(2^{y-1})*(5^{x-1}), what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

E

Protocol:Simplify expression

1. need to find X and Y but cant isolate X and Y directly so start by separating the bases:

divide both sides by (5^{x-1}

left and right side simplifies to 5-5^{y-x+1} = 2^{y-1}

2. after simplifying, analyze.

Right side must be positive integer ( y is at least 1 ) thus left side must be positive too.
==> Left side is 5 minus an expression so answer must be at max 4 and at minimum 1. Right side must also be equal to a power of 2. Thus 4 is only possible answer for left side. Thus Y = 3.

Note: From original expression it is clear that X> Y. However by the time you get to simplifying the expression, the possible number of answers is so constrained that this information isn't critical to arriving at a faster answer.
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Re: If x and y are positive integers and 5^x  [#permalink]

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01 Aug 2017, 09:57
1
Top Contributor
BN1989 wrote:
If x and y are positive integers and $$(5^x)-(5^y)=(2^{y-1})*(5^{x-1})$$, what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

Given: $$(5^x)-(5^y)=(2^{y-1})*(5^{x-1})$$

Divide both sides by $$5^{x-1}$$ to get: $$5^1 - 5^{y-x+1} = 2^{y-1}$$

Simplify: $$5 - 5^{y-x+1} = 2^{y-1}$$

OBSERVE: Notice that the right side, $$2^{y-1}$$, is POSITIVE for all values of y

Since y is a positive integer, $$2^{y-1}$$ can equal 1, 2, 4, 8, 16 etc (powers of 2)

So, the left side, $$5 - 5^{y-x+1}$$, must be equal 1, 2, 4, 8, 16 etc (powers of 2).

Since $$5^{y-x+1}$$ is always positive, we can see that $$5 - 5^{y-x+1}$$ cannot be greater than 5

So, the only possible values of $$5 - 5^{y-x+1}$$ are 1, 2 or 4

In other words, it must be the case that:
case a) $$5 - 5^{y-x+1} = 2^{y-1} = 1$$
case b) $$5 - 5^{y-x+1} = 2^{y-1} = 2$$
case c) $$5 - 5^{y-x+1} = 2^{y-1} = 4$$

Let's test all 3 options.

case a) $$5 - 5^{y-x+1} = 2^{y-1} = 1$$
This means y = 1 (so that the right side evaluates to 1)
The left side, $$5 - 5^{y-x+1} = 1$$, when $$5^{y-x+1} = 4$$. Since x and y are positive integers, it's IMPOSSIBLE for $$5^{y-x+1}$$ to equal 4
So, we can eliminate case a

case b) $$5 - 5^{y-x+1} = 2^{y-1} = 2$$
This means y = 2 (so that the right side evaluates to 2)
The left side, $$5 - 5^{y-x+1} = 2$$, when $$5^{y-x+1} = 3$$. Since x and y are positive integers, it's IMPOSSIBLE for $$5^{y-x+1}$$ to equal 3
So, we can eliminate case b

case c) $$5 - 5^{y-x+1} = 2^{y-1} = 4$$
This means y = 3 (so that the right side evaluates to 4)
The left side, $$5 - 5^{y-x+1} = 4$$, when $$5^{y-x+1} = 1$$.
If $$5^{y-x+1} = 1$$, then $$y-x+1 = 0$$
In this case, y = 3
So, we can write: 3 - x + 1 = 0, which mean x = 4
So, the only possible solution is y = 3 and x = 4, which means xy = (4)(3) = 12

Cheers,
Brent
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Re: If x and y are positive integers and 5^x  [#permalink]

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07 Aug 2017, 22:37
1
BN1989 wrote:
If x and y are positive integers and $$(5^x)-(5^y)=(2^{y-1})*(5^{x-1})$$, what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

$$(5^x)-(5^y)=(2^{y-1})*(5^{x-1})$$
=$$(5^{x-1})(5-5^{y-x+1})=(2^{y-1})*(5^{x-1})$$

5^{x-1} =/=0. So, $$(5-5^{y-x+1})=(2^{y-1})$$
Since $$(2^{y-1})$$ must be +ve. Also Y is +ve so, y-1>0 and hence $$(2^{y-1})$$ will be an integer only.
Hence y-x+1 can be 0 only.
y-x+1 = 0
-> y-x = -1

also at y-x+1 = 0
$$5-1 = (2^{y-1})$$

$$4 = (2^{y-1})$$
y -1 = 2
y = 3

x= y+1 = 4
xy = 12

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If x and y are positive integers and 5^x  [#permalink]

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02 Jul 2018, 16:03
1
BN1989 wrote:
If x and y are positive integers and $$(5^x)-(5^y)=(2^{y-1})*(5^{x-1})$$, what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

$$5^x-5^y=2^{y-1}*5^{x-1}$$

$$\frac{5^x}{5^{x-1}}-\frac{5^y}{5^{x-1}}=2^{y-1}$$

$$5-5^{y-x+1}=2^{y-1}$$

The blue equation implies the following:
5 - POWER OF 5 = POWER OF 2.
The only logical option is as follows:
$$5 - 5^0 = 2^2$$.

Since the right side of the blue equation is equal to $$2^2$$, we get:
$$2^{y-1}=2^2$$
$$y-1=2$$
$$y=3$$.

Since y=3 and the subtracted term on the left side is equal to $$5^0$$, we get:
$$5^{3-x+1}=5^0$$
$$3-x+1=0$$
$$4=x$$.
.
Thus:
$$xy = 4*3 = 12$$.

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Re: If x and y are positive integers and 5^x  [#permalink]

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10 Aug 2019, 09:00
1
Clearly,
5^x - 5^y = 2^y-1 * 5^x-1
(5^x-5^y)/5^(x-1) = 2^(y-1)
5 - 5^(y-x+1) = 2^ (y-1)

Now,keeping conditions RHS always +ve .So LHS must always be a positive.
RHS , a factor of 2 can be made on the LHS by when 5^(y-x+1) = 1
So y-x+1 = 0
y-x=-1

Also y-1 =2
so y =3
x =4

which gives xy = 12.
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Re: If x and y are positive integers and 5^x  [#permalink]

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24 Sep 2013, 15:38
Bunuel wrote:
BN1989 wrote:
If x and y are positive integers and (5^x)-(5^y)=(2^(y-1))*(5^(x-1)), what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

Notice that we are told that $$x$$ and $$y$$ are positive integers.

$$5^x-5^y=2^{y-1}*5^{x-1}$$ --> $$5^x-2^{y-1}*5^{x-1}=5^y$$ --> $$5^x(1-\frac{2^y}{2}*\frac{1}{5})=5^y$$ --> $$5^x(10-2^y)=2*5^{y+1}$$.

Now, since the right hand side is always positive then the left hand side must also be positive, hence $$10-2^y$$ must be positive, which means that $$y$$ can take only 3 values: 1, 2 and 3.

By trial and error we can find that only $$y=3$$ gives integer value for $$x$$: $$5^x(10-2^3)=2*5^{3+1}$$ --> $$2*5^x=2*5^4$$ --> $$x=4$$ --> $$xy=12$$.

Bunuel, I have a question, how did you know that you had to reorganize the equation in this way?

From $$5^x-5^y=2^{y-1}*5^{x-1}$$ TO $$5^x-2^{y-1}*5^{x-1}=5^y$$

Thanks!
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If x and y are positive integers and 5^x  [#permalink]

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19 Aug 2015, 07:30
Bunuel wrote:
BN1989 wrote:
If x and y are positive integers and (5^x)-(5^y)=(2^(y-1))*(5^(x-1)), what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

Notice that we are told that $$x$$ and $$y$$ are positive integers.

$$5^x-5^y=2^{y-1}*5^{x-1}$$;

$$5^x-2^{y-1}*5^{x-1}=5^y$$;

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

$$5^x(10-2^y)=2*5^{y+1}$$.

Now, since the right hand side is always positive then the left hand side must also be positive, hence $$10-2^y$$ must be positive, which means that $$y$$ can take only 3 values: 1, 2 and 3.

By trial and error we can find that only $$y=3$$ gives integer value for $$x$$:

$$5^x(10-2^3)=2*5^{3+1}$$;

$$2*5^x=2*5^4$$;

$$x=4$$ --> $$xy=12$$.

i did another logic is it right?

$$(5^x)(1-(2^y-1)/5)=5^y$$

then $$(5^x-y-1)(5-(2^y-1))=1$$

which means this must be $$x-y-1= 0$$
and $$5-(2^y-1) = 1$$ means also $$2^y-1 = 4$$ then $$y-1=2$$
then y=3 replace y in old equation we get x =4 then finally xy =12
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Re: If x and y are positive integers and 5^x  [#permalink]

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19 Aug 2015, 08:04
(5^x)-(5^y)=(2^(y-1))*(5^(x-1))
(5^x)(1-5^y-x) = (2^(y-1))*(5^(x-1))
5(1-5^y-x) = (2^(y-1))
10(1-5^y-x) = (2^(y))

We note that (2^(y)) is always positive. which translates to (1-5^y-x) >0 or y-x<0

So, 10(5^y-x)(5^(x-y)-1) = (2^(y))
5^(y-x+1)*2*(5^(x-y)-1) =(2^(y))

Now, RHS is 5^0, which means y-x+1 = 0, x-y =1

inputting values in above,

5^(0)*2*(5^(1)-1) =(2^(y))
2*4 = (2^(y))
implies y =3
x= 4

xy = 12
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Re: If x and y are positive integers and 5^x  [#permalink]

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21 Sep 2016, 02:31
Hi,
The explanations are great. However, I took a lot of time figuring out how to simplify the equation and get to some solution and then started exploring otherway around which worked faster for me..

Difference between any 2 powers of 5 would always yields an even number.,i.e., a multiple of 2. However, difference of only consecutive powers of 5 yields a number that is only multiple of 2 and 5. Check->(25-5), (125-25), (625-125). Also check(625-25), (625-5), etc. Then, I simply had to check the options which had consecutive integers as factors. Only 12 worked out with 3 and 4 as factors.

This is not a foolproof solution but just another way of thinking incase you feel trapped in a question.
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Re: If x and y are positive integers and 5^x  [#permalink]

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30 Jan 2018, 14:30
Hi All,

GMAT Quant questions can almost always be solved in a variety of ways, so if you find yourself not able to solve by doing complex-looking math, then you should look for other ways to get to the answer. Think about what's in the question; think about how the rules of math "work." This question is LOADED with Number Property clues - when combined with a bit of "brute force", you can answer this question by doing a lot of little steps.

Here are the Number Properties (and the deductions you can make as you work through them):

1) We're told that X and Y are POSITIVE INTEGERS, which is a great "restriction."
2) We can calculate powers of 5 and powers of 2 rather easily:

5^0 = 1
5^1 = 5
5^2 = 25
5^3 = 125
5^4 = 625
Etc.

2^0 = 1
2^1 = 2
2^2 = 4
Etc.

3) The answer choices are ALL multiples of 3. Since we're asked for the value of XY, either X or Y (or both) MUST be a multiple of 3.

4)
*The left side of the equation is a positive number MINUS a positive number.
*The right side is the PRODUCT of two positive numbers, which is POSITIVE.
*This means that 5^X > 5^Y, so X > Y.

5)
*Notice how we have 5^X (on the left side) and 5^(X-1) on the right side; these are consecutive powers of 5, so the first is 5 TIMES bigger than the second.
*On the left, we're subtracting a number from 5^X.
*On the right, we're multiplying 2^(Y-1) times 5^(X-1).
*2^(Y-1) has to equal 1, 2 or 4, since if it were any bigger, then multiplying by that value would make the right side of the equation TOO BIG (you'd have a product that was bigger than 5^X).
*By extension, Y MUST equal 1, 2 or 3. It CANNOT be anything else.

6) Remember that at least one of the variables had to be a multiple of 3. What if Y = 3? Let's see what happens….

5^X - 5^3 = 2^2(5^(X-1))

5^X - 125 = 4(5^(X-1))

Remember that X > Y, so what if X = 4?…..

5^4 - 125 = 500
4(5^3) = 500

The values MATCH. This means Y = 3 and X = 4. XY = 12

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Re: If x and y are positive integers and 5^x  [#permalink]

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31 Jan 2018, 02:10
BN1989 wrote:
If x and y are positive integers and $$(5^x)-(5^y)=(2^{y-1})*(5^{x-1})$$, what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

Good Question.. I solved it this way..

Let $$x-1 = a$$, $$y-1 = b$$.. So the equation becomes,

$$5^a - 5^b$$ = $$\frac{5^a*2^b}{5}..$$
Now, we have to find the value of $$xy$$ i.e. $$(a+1)(b+1)$$
By looking at the options and the above equation only E satisfies.
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Re: If x and y are positive integers and 5^x  [#permalink]

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02 Feb 2018, 12:09
BN1989 wrote:
If x and y are positive integers and $$(5^x)-(5^y)=(2^{y-1})*(5^{x-1})$$, what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

We can simplify the given equation by dividing both sides by 5^(x - 1):

5^x/5^(x - 1) - 5^y/5^(x - 1) = 2^(y-1)

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

It’s not easy to solve an equation with two variables by algebraic means; however, since both variables are positive integers, we can try numbers for one of the variables and solve for the other.

If we let y = 1, then the right hand side (RHS) = 2^0 = 1 and thus the left hand side (LHS) is 5^(1 - x + 1) = 4. However, a power of 5 can’t be equal to 4 when the exponent is an integer.

Now, let’s let y = 2; then the RHS = 2^1 = 2 and thus, for the LHS, 5^(2 - x + 1) = 3. However, a power of 5 can’t be equal to 3 when the exponent is an integer.

Finally, let’s let y = 3; then the RHS = 2^2 = 4 and thus, for the LHS, 5^(3 - x + 1) = 1. We see that a power of 5 can be equal to 1 when the exponent is 0. Thus:

3 - x + 1 = 0

x = 4

We see that x = 4 and y = 3 and thus xy = 12.

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If x and y are positive integers and 5^x  [#permalink]

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02 Jul 2018, 13:15
-- The simplest solution so far --

1/ Move the 5^(x-1) to the left side
2/ Move (1/2) from the right side 2^(y-1) = (2^y)*(1/2) to the left side
3/ Now you have 10 - 2*5^(y-x+1) = 2^y
4/ Ask when this is possible? -- The right side is always >0 so the left side must be >0 and this is true only if (y-x+1)=0
5/ Now you are looking for a combination of numbers from the choices that is y+1=x
6/ 3*4=12 looks good!
If x and y are positive integers and 5^x   [#permalink] 02 Jul 2018, 13:15

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