SOLUTION: {{{ 2^(x+1) + 2^x = 3^(y+2) - 3^y }}}, where x,y ϵ ℤ. Find x.

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: {{{ 2^(x+1) + 2^x = 3^(y+2) - 3^y }}}, where x,y ϵ ℤ. Find x.      Log On


   

Question 1191422: +2%5E%28x%2B1%29+%2B+2%5Ex+=+3%5E%28y%2B2%29+-+3%5Ey+, where x,y ϵ ℤ. Find x.
Found 2 solutions by Alan3354, math_tutor2020:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
+2%5E%28x%2B1%29+%2B+2%5Ex+=+3%5E%28y%2B2%29+-+3%5Ey+, where x,y ϵ ℤ. Find x.
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x = 3

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

+2%5E%28x%2B1%29+%2B+2%5Ex+=+3%5E%28y%2B2%29+-+3%5Ey+

+2%5Ex%2A2%5E1+%2B+2%5Ex+=+3%5Ey%2A3%5E2+-+3%5Ey+

+2%2A2%5Ex+%2B+2%5Ex+=+9%2A3%5Ey+-+3%5Ey+

+3%2A2%5Ex+=+8%2A3%5Ey+

+3%2A2%5Ex+=+2%5E3%2A3%5Ey+

+%282%5Ex%29%2F%282%5E3%29+=+%283%5Ey%29%2F%283%5E1%29+

+2%5E%28x-3%29+=+3%5E%28y-1%29+

If we make both exponents to be zero, then we must have x = 3 and y = 1

+2%5E%283-3%29+=+3%5E%281-1%29+

+2%5E0+=+3%5E0+

1+=+1

If y+%3C%3E+0, then we can't have x = 3 or else the sides won't match up. The same goes for when x+%3C%3E+0 and we try to find y values that might work.

The left hand side represents powers of 2, and the right hand side represents powers of 3. The two primes don't match up to line up anywhere else except for when 2^0 = 3^0 = 1.

Otherwise, 2%5EA+=+3%5EB doesn't have any integer solutions for nonzero integers A,B.

Also, the right hand side grows faster than the left side. This means the two curves only intersect when (x,y) = (3,1).

Answer: x = 3