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)# ?

1 Answer
Apr 14, 2017

Answer:

#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#