log_y(x) + log_x(y) = 43

What are the values of x and y?

This one doesn't appear to have any nice solutions, but here's how I think I solved it.

First, let's say logx(y) = a. This means logy(x) = 1/a.

(Reason:
Assume logx(y)=a and logy(x)=b;
Then, (x^a)=y and (y^b)=x;
Substituting gives ((x^a)^b)=x;
x^a = x^(1/b);
a = 1/b;
Thus, logx(y) = 1/logy(x);
)

So, a + 1/a = 43. This is equivalent to a^2 + 1 = 43*a, or a^2 - 43*a + 1 = 0.
Solving for "a" yields either 43.977... or 0.023...
These are recipricals of each other since x and y are interchangable.

Thus, the solution is the set of numbers x^42.977... = y, where x and y are not 0 or 1. I cannot find an integer solution to this. One approximate solution is x = 2, y = 8,656,974,208,780.

Posted by Ender on 2002-06-04 07:01:02