Question

How do you solve the system of equations: `A^(A+1)+B^(B+1)+C^(C+1) = 729` and `A^(1/A) = B^(1/B) = C^(1/C)` ?

Answer 1

`A^(1/A) ~~ 1.432024273098`

Explanation:

The simplest conditions under which:

`A^(1/A) = B^(1/B) = C^(1/C)`

are when `A=B=C`

There are other solutions, but let's stick with that for now.

In which case:

`3*A^A A = 729`

So:

`A^(A+1)= A^A A = 729/3 = 243 = 3^5`

If the question had specified `243 = 3^5` instead of `729=3^6` then we would find `A=3` and `A^(1/A) = root(3)(3)`. I suspect that this is what the question should have been and `729` is in error.

As it is, the solution is not expressible in terms of elementary functions. We can find approximations using Newton's method.

Let:

`f(x) = x^(x+1) - 243`

Then:

`f'(x) = (x+1)x^x+x^(x+1)ln x`

by Newton's method, if `a_i` is an approximation to a zero of `f(x)`, then a better approximation is given by:

`a_(i+1) = a_i - (f(a_i))/(f'(a_i))`

So in our example, we would probably choose `a_0 = 3` and iterate using the formula:

`a_(i+1) = a_i - (a_i^(a_i+1) - 243)/((a_i+1)a_i^(a_i)+a_i^(a_i+1)ln a_i)`

Putting this in a spreadsheet and iterating, I found:

`a_0 = 3`

`a_1 ~~ 3.822386809107`

`a_2 ~~ 3.583342155402`

`a_3 ~~ 3.465951810855`

`a_4 ~~ 3.443670010969`

`a_5 ~~ 3.442996111113`

`a_6 ~~ 3.442995518121`

`a_7 ~~ 3.442995518121`

Using `A=3.442995518121`, we find:

`A^(1/A) ~~ 1.432024273098`