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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.

## Hi Brent, the answer given to

https://gmatclub.com/forum/is-4-x-y-139120.html

Is 4^(x+y) = 8^10?

(1) x - y = 9

(2) y/x = 1/4

The equation becomes: x+y = 15

We can solve both the equation with both the statements. Therefore both are sufficient for me.

## Be careful. You have

Be careful. You have accidentally turned the target QUESTION into a true STATEMENT.

The target question asks "Is 4^(x+y) = 8^10?"

We can rephrase this to get: "Does x + y = 15?"

We still have a QUESTION on our hands. We don't know whether or not x + y = 15. Our goal is to determine whether each statement provides enough information to answer that question.

Statement 1) x - y = 9

Does this statement provide sufficient information to answer the rephrased target question ("Does x + y = 15?")?

No.

If x - y = 9, then it's possible that x = 12 and y = 3, in which case the answer to our rephrased target question is "YES, x + y does equal 15

If x - y = 9, then it's also possible that x = 10 and y = 1, in which case the answer to our rephrased target question is "NO, x + y does NOT equal 15

So, statement 1 is not sufficient.

Does that help?

By the way, here's my step-by-step solution: https://gmatclub.com/forum/is-4-x-y-139120-20.html#p1901036

Cheers,

Brent

## Hi Brent, could you please

https://gmatclub.com/forum/x-is-an-integer-and-x-raised-to-any-odd-integer-is-greater-95476.html

## Question link: https:/

Question link: https://gmatclub.com/forum/x-is-an-integer-and-x-raised-to-any-odd-integ...

VERY TRICKY!!!

Here's my solution: https://gmatclub.com/forum/x-is-an-integer-and-x-raised-to-any-odd-integ...

Cheers,

Brent

## https://gmatclub.com/forum/x

can you please help me with an explanation.

## You bet.

You bet.

Here's my full solution: https://gmatclub.com/forum/x-is-an-integer-and-x-raised-to-any-odd-integ...

Cheers,

Brent

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