good exponential question

rajeshvellanki · Posted: Fri Apr 20, 2007 9:41 am Post subject: good exponential question

x and y are positive integers 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

Answer E

(5^x-5^y)/5^(x-1)=2^(x-1)

5-5^(y-x+1)=2^(y-1)

5=2^(y-1)+5^(y-x+1)

y-x+1=0 for 5=2^(y-1)+5^(y-x+1)

and 2^(y-1)=4 ==> y=3 and x=4

xy=12

answer E

GMATpaduan · Posted: Tue Jun 12, 2007 9:01 am Post subject: Please explain

Rajesh - You lost me a little bit with your explanation. Can someone provide more detail please?

I have everything down except how you go from-----

5=2^(y-1)+5^(y-x+1)

To: ----------------

y-x+1=0 for 5=2^(y-1)+5^(y-x+1)

and 2^(y-1)=4 ==> y=3 and x=4

xy=12

answer E

How do you know y-x+1 = 0? and that 2^(y-1) = 4?

induscrede · Posted: Sun Jul 08, 2007 2:05 pm Post subject:

Closer examination will reveal that this is the only way the equation can equal 5

5^(y-x+1) = 5^0= 1 (has to)so that we can get 1 for the latter part of the equation, anything more that that will equal 5 and up and .
In case of (y-x+1) = 1, there is no value for 2^something to equal 0 , for the tntire equation to equal 5

2^0=1 (Anything to the power zero is just "1"), which will make the equation =6, so we have to have the equation break down as 5= 4+1.

Hope that helps explaining the "Jump"

iomail · Posted: Fri Jul 13, 2007 2:55 pm Post subject:

(5^x-5^y)/5^(x-1)=2^(x-1) I got lost too...
first: shouldn't the latter part of the equation above be ...=2^(y-1) ??

5-5^(y-x+1)=2^(y-1)
and then here above, once you put together the second part of the first equation, shouldn't we have 5^x-5^(y-x+1)=...???
Otherwise, I didn't understand the jump!

5=2^(y-1)+5^(y-x+1)

y-x+1=0 for 5=2^(y-1)+5^(y-x+1)

and 2^(y-1)=4 ==> y=3 and x=4

xy=12

answer E