Lesson: Equations and Powers

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I need to understand the logic of this problem: How do we know what "x" equals and what "y" equals?

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
gmat-admin's picture

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 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

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???
gmat-admin's picture

Yes, that's perfect.
It's a super tough question to answer quickly.

Hi Brent, the answer given to the question below is c. I am failing to understand why?
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.
gmat-admin's picture

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 help me with an explanation to this question below?

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

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

can you please help me with an explanation.
gmat-admin's picture

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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