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
good exponential question
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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?
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?
GOOD LUCK ALL!
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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"
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"
(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
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