I happen to come across this question in one of the forums..I would certainly appreciate if anyone could provide the solution along with a detailed explanation
5^x-5^y=2^(y-1)*5^(x-1).What is the value of xy?
I don't have the answer choices with me...M sorry about that.
Can you solve this one
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 133
- Joined: Sat Dec 29, 2007 2:43 am
- Thanked: 12 times
-
- Master | Next Rank: 500 Posts
- Posts: 379
- Joined: Wed Jun 03, 2009 3:05 am
- Thanked: 19 times
- Followed by:1 members
- GMAT Score:690
5^x - 5^y = 2^(y-1)*5^(x-1)
=> (5^x - 5^y)/(5^(x-1)) = 2^(y-1)
=> 5 - 5^(y+1-x) = 2^(y-1)
=> 5 = 5^(y+1-x) + 2^(y-1)
Now, a sum of 5 can be achieved with two terms as (2+3 or 3+2 or 1+4 or 4+1).
Of the above, only 1+4 seems feasible.
Hence, 5^(y+1-x) = 1 and 2^(y -1) = 4 = 2^2
=> y+1-x = 0 and y-1 =2 => y = 3
=> 3+1-x = 0
=> x=4
=> xy = 12.
=> (5^x - 5^y)/(5^(x-1)) = 2^(y-1)
=> 5 - 5^(y+1-x) = 2^(y-1)
=> 5 = 5^(y+1-x) + 2^(y-1)
Now, a sum of 5 can be achieved with two terms as (2+3 or 3+2 or 1+4 or 4+1).
Of the above, only 1+4 seems feasible.
Hence, 5^(y+1-x) = 1 and 2^(y -1) = 4 = 2^2
=> y+1-x = 0 and y-1 =2 => y = 3
=> 3+1-x = 0
=> x=4
=> xy = 12.
-
- Junior | Next Rank: 30 Posts
- Posts: 10
- Joined: Wed Dec 16, 2009 6:39 am
Hi sreak 1089
Can you please explain how did you get from the second to the third line.
(5^x - 5^y)/(5^(x-1)) = 2^(y-1)
=> 5 - 5^(y+1-x) = 2^(y-1)
Can you please explain how did you get from the second to the third line.
(5^x - 5^y)/(5^(x-1)) = 2^(y-1)
=> 5 - 5^(y+1-x) = 2^(y-1)
-
- Master | Next Rank: 500 Posts
- Posts: 379
- Joined: Wed Jun 03, 2009 3:05 am
- Thanked: 19 times
- Followed by:1 members
- GMAT Score:690
Certainly.
(5^x - 5^y)/(5^(x-1)) can be written as:
(5^x /(5^(x-1)) - (5^y)/(5^(x-1))
Bring the exponents in the denominator to numerator:
=> (5^(x-(x-1)) - (5^(y-(x-1))
=> (5^(x-x+1)) - (5^(y-x+1))
=> 5 - (5^(y-x+1))
Hope that helps.....
(5^x - 5^y)/(5^(x-1)) can be written as:
(5^x /(5^(x-1)) - (5^y)/(5^(x-1))
Bring the exponents in the denominator to numerator:
=> (5^(x-(x-1)) - (5^(y-(x-1))
=> (5^(x-x+1)) - (5^(y-x+1))
=> 5 - (5^(y-x+1))
Hope that helps.....
imanemekouar wrote:Hi sreak 1089
Can you please explain how did you get from the second to the third line.
(5^x - 5^y)/(5^(x-1)) = 2^(y-1)
=> 5 - 5^(y+1-x) = 2^(y-1)
-
- Junior | Next Rank: 30 Posts
- Posts: 10
- Joined: Wed Dec 16, 2009 6:39 am
thank you for your quick response cherck. But Wondering , how did you conclude that 5^Y-X+1 have to be equal to 1.
-
- Master | Next Rank: 500 Posts
- Posts: 379
- Joined: Wed Jun 03, 2009 3:05 am
- Thanked: 19 times
- Followed by:1 members
- GMAT Score:690
We know that the sum of the two terms is 5, which can be 2+3, 3+2 or 1+4 or 4+1. Now, we have one term a power of 2 and the other a power of 5. Now power 5 can not be 2, 3 or 4, so it has to be one only.
imanemekouar wrote:thank you for your quick response cherck. But Wondering , how did you conclude that 5^Y-X+1 have to be equal to 1.
-
- Junior | Next Rank: 30 Posts
- Posts: 10
- Joined: Wed Dec 16, 2009 6:39 am