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## Comment on

Equations and Powers## I need to understand the

Here is what I am talking about:

OBSERVE: Notice that the right side, 2^(y−1), is POSITIVE for all values of y -----OK

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

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

Since 5^(y−x+1) is always positive, we can see that 5 − 5^(y−x+1) cannot be greater than 5 ----so what are the values for x and y that I plug in ?

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

## You're referring to my

You're referring to my solution to this question: https://gmatclub.com/forum/if-x-and-y-are-positive-integers-and-5-x-1303...

We're not plugging in; we're solving equations.

Once we make all of the above conclusions, we know that there are only 3 cases (as shown in my solution)

Take case a for example: 5 − 5^(y−x+1) = 2^(y−1) = 1

If 2^(y−1) = 1, then y = 1

If 5 − 5^(y−x+1) = 1, then 5^(y−x+1) = 4

Since it's IMPOSSIBLE for 5^(some integer) to equal 4, we can eliminate case a.

Then we move onto case b...etc

For each case, we're solving an equation. No plugging in necessary.

## You're referring to my

We're not plugging in; we're solving equations.

Once we make all of the above conclusions, we know that there are only 3 cases (as shown in my solution)

Take case “a” for example: 5 − 5^(y−x+1) = 2^(y−1) = 1

If 2^(y−1) = 1, then y = 1

Since y is the first multiple of 2(and since it has to be positive) we plug it in 2^(1−1) = 20 =1

OK

If 5 − 5^(y−x+1) = 1, then 5^(y−x+1) = 4

1) 5 − 5^(y−x+1) = 1

2) − 5^(y−x+1) = -4

3) (-1)^( − 5^(y−x+1) = -4)

4) 5^(y−x+1) = 4 (5 to any power cannot equal 4, so no good?)

So now I am assuming we use the next power of 2 (which is 2) and plug it in?

1) 5 − 5^(y−x+1) = 2

2) − 5^(y−x+1) = -5 + (2)

3) (-1)( − 5^(y−x+1) = -3 )

4) 5^(y−x+1) = 3 (5 to any power cannot equal 3, so no good?)

On to the next power of 2 (which is 4) and plug it in?

1) 5 − 5^(y−x+1) = 4

2) − 5^(y−x+1) = -5 + (4)

3) (-1)( − 5^(y−x+1) = -1)

4) 5^(y−x+1) = 1 (5 to “0” can equal 4, so good, right?)

Did I approach this the right way?

I don’t know how you would do this problem in under 2 minutes???

## Yes, that's perfect.

Yes, that's perfect.

It's a super tough question to answer quickly.

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