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.

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)

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.....

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)

thank you for your quick response cherck. But Wondering , how did you conclude that 5^Y-X+1 have to be equal to 1.

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.

Tkans you.