Hello, jmedsy!

A fascinating problem . . .

Solve for

. . .

. .

Take logs of equation [1]: .

Take logs of equation [2]: .

We have a system of equations: .

Substitute: .

From [3], we have: .

Substitute into [4]: .

Factor: .

. .

. .

We have: .

Back-substitute:

. .

. .

Two solutions: .