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 =
3So, we can write:
3 - x + 1 = 0, which mean
x = 4So, the only possible solution is
y = 3 and
x = 4, which means xy = (
4)(
3) = 12
Cheers,
Brent
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