If x and y are integers and ...

[Joanna]

Hi, I have a question about a particular step in the simplification of this exponential equation:

Full question in CAT:

If x and y are integers and

(15^x + 15^x+1)/4^y=15^y

what is the value of x?
2
3
4
5
cannot be determined

ALL VARIABLES ARE ACTUALLY EXPONENTS
My question is how do you get from the first red equation to the second red.
--------------------------------------------------------------
(15x + 15x+1) = 15y4y
[15x + 15x(15^1)] = 15y4y
(15x )(1 + 15) = 15y4y
(15x)(16) = 15y4y
(3x)(5x)(24) = (3y)(5y)(22y)

[esledge]

At that step, we are factoring out a 15^x. It might help if I show a couple intermediate steps (also in red):

15^x + (15^x)(15^1) = (15^y)(4^y)
(15^x)(1) + (15^x)(15^1) = (15^y)(4^y)
(15^x)(1 + 15^1) = (15^y)(4^y)
(15^x)(1 + 15) = (15^y)(4^y)
(15^x)(16) = (15^y)(4^y)

From the (15^x)(1 + 15^1) step, you might also verify the logic by distributing the 15^x back over the stuff in parentheses: (15^x)(1) + (15^x)(15^1). This checks out with the previous line in the manipulation.

[Matt]

obviously from the way it's been broken down in the posts above, x = 4 and y = 4 is a solution, but is it possible that there are others? what is the OA?

[MICHAEL_SHAUNN]

Well I believe that the answer is not 4.
Actually the answer is x=2 and y=2.Here's how I reached to my answer:

(15^x)(16)=(4^y)(15^y)
(15^x)(4^2)=(4^y)(15^y)
Comparing the powers,we get y=2 and x=y i.e. x=y=2.

Let's look at it in another way:

(15^x)(4^2)=(4^y)(15^y)
(15^x)/(15^y)=(4^y)/(4^2)
15^(x-y)=4^(y-2)
now,15 raised to any power will always have the one's digit as 5 and 4 raised to any power will have the one's digit as either 4 or 6.(if x>y and y>2 blah blah blah....as they wont make any differnence to the logic).Hence the only solution will exist if both sides are equal to 1 i.e x=y and y=2 i.e. x=y=2.

THANKS!!

[esledge]

Michael is correct: x = y = 2. That units digit approach is neat. I personally would solve it the first way (setting the exponents equal once you spot the equal bases), but it's nice to look at it a different way. Thanks!