log
y(x) + log
x(y) = 43
What are the values of x and y?
One equation, two unknowns. In general, that means an infinite number of solutions. I'll generate a couple (forgive the lack of subscripts):
logx(y) = logy(y)/logy(x) = 1/logy(x)
So from the original problem,
logy(x) + 1/logy(x) = 43
Let L = logy(x).
L + 1/L = 43
L^2 - 43L + 1 = 0
L = (43 +/- sqrt(43^2 - 4))/2
L has roots at 42.9767 and 0.02327, so logy(x) = 42.9767 or logy(x) = 0.02327. Either one will work (Note that these roots are reciprocals of one another).
One solution would be (2, 2^42.9767). Another would be (10, 10^42.9767). Still another would be (2^42.9767, 2). You get the idea.