Indices and Surds?

Question

Solve this if you can:
Let x = (1 + 1/2002)^2002 and let y = (1 + 1/2002)^2003

Find (x^y)/(y^x)

Sorry guys, I forget to add that not calculator is allowed when solving this question.
p.s. (the correct answer is 1)

First to solve it gets my best answer vote!

Answer 1

ok:

Let 2002=b and let 1/2002=1/b=a

Then x=(1+a)^b, and y=(1+a)^(b+1)

Therefore x^y= [(1+a)^b]^[(1+a)^(b+1)] = (1+a)^[b • (1+a)^b • (1+a)]

but a=1/b so b(1+a) = b(1+1/b) = b+1

Thus x^y = (1+a)^[(b+1) • (1+a)^b]


and y^x = [(1+a)^(b+1)]^[(1+a)^b] = (1+a)^[(b+1) • (1+a)^b] = x^y


Thus (x^y)/(y^x)=1

Even the rounding error in Maple is too big to give the correct answer . . .

Answer 2

Let z=(1+1/2002), Then in terms of z we have:

x=z^2002
y=z^2003=(z^2002)*z=x*z

So we have: (x^y)/(y^x) =
(x^(xz))/((x*z)^x))
=(x^(x*(z-1))/(z^x)

but z-1=1/2002, so we have:
(z^(2002*x/2002))/(z^x) [ using x=z^2002
=z^x/z^x
=1

Answer 3

Let me try.

1 + 1/2002 = 1.0004995004

Therefore x = 2.94391

Similarly, y = 2.71896

x^y = 18.83592

y^x = 19.00390

Now, (x^y) (y^x) = 18.83592 * 19.00390

= 357.9559

Therefore the answer is 357.9559. If it is rounded then the answer is 358.