exponents

[willbeatthegmat]

x & y are positive integers.If 5^x - 5^y= 2^(y-1).5^(x-1), wat is d value of xy?

48
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18
12




ans 12

[DanaJ]

I'll assume that point = multiplication or *
5^x - 5^y= 2^(y-1)*5^(x-1) is equivalent to:
5^x - 2^(y-1)*5^(x - 1) = 5^y (i just switched the places between the two).
Factor 5^(x - 1) and you get that
5^(x - 1)*[5 - 2^(y - 1)] = 5^y.
Now, since 5^(x - 1) and 5^y are both positive (they are powers of a positive number - in this case 5), then 5 - 2^(y - 1) must be positive as well. This means in turn that 5 > 2^(y - 1) and there are only 3 cases for this one:
a. y = 1, with 2^(y - 1) = 1, and with 5 - 2^(y - 1) = 4. But this one is not right, since (5 to a power)*4 cannot equal (5 to a different power).
b. y = 2, with 2^(y - 1) = 2 and 5 - 2^(y - 1) = 3. Again, this doesn't work, since that 3 cannot "come between" powers of 5.
c. y = 3, with 2^(y - 1) = 4 and 5 - 2^(y - 1) = 1. This is the case we're looking for, since we get 5^(x - 1)*1 = 5^3. This means that x - 1 = 3, with x = 4.
This gives us xy = 12.

[cjb]

x & y are positive integers.If 5^x - 5^y= 2^(y-1).5^(x-1), wat is d value of xy?

48
36
24
18
12

ans 12

Stared at this one for a while, and worked some things out, like x > y, but in the end I just picked numbers.

I started with x=3, y=2, making the LHS 125-25 = 100. That doesn't work, but the next pair of powers does:
625-125=500
x=4, y=3

So the answer is E.

Notice that the first pair of numbers I picked wasn't a possible answer - I should have saved myself the time.

[logitech]

So since the right side of the equation has 2 there is only one way to get it using factors of 5:

5-1 = 4


In other words:

5^y [( 5^(x-y) - 1)] = 5^(x-1) 2^(y-1)

so we can match the 5s and 2s:

y=x-1 , which gives us x=y+1

if we use this at:

[( 5^(x-y) - 1)] = [5^(y-y-1) -1] = 5-1 = 4

So:

2^(y-1) = 4

y=3 and x=4 ; xy=12

[kanha81]

So since the right side of the equation has 2 there is only one way to get it using factors of 5:

5-1 = 4


In other words:

5^y [( 5^(x-y) - 1)] = 5^(x-1) 2^(y-1)

so we can match the 5s and 2s:

y=x-1 , which gives us x=y+1

if we use this at:

[( 5^(x-y) - 1)] = [5^(y-y-1) -1] = 5-1 = 4

So:

2^(y-1) = 4

y=3 and x=4 ; xy=12

Wow! This is quite something beautiful.

[aj5105]

Can somebody explain the below solution? I got stuck in the middle.

Thanks.
AJ

So since the right side of the equation has 2 there is only one way to get it using factors of 5:

5-1 = 4


In other words:

5^y [( 5^(x-y) - 1)] = 5^(x-1) 2^(y-1)

so we can match the 5s and 2s:

y=x-1 , which gives us x=y+1

if we use this at:

[( 5^(x-y) - 1)] = [5^(y-y-1) -1] = 5-1 = 4

So:

2^(y-1) = 4

y=3 and x=4 ; xy=12