I stumbled across the answer. I re-arranged the original equation to:
[x-1]^(x^2/2) = [1/x]^(x + 1 - x^2/2)
Looked at that for a long time going nowhere.
Then I started thinking = bases and = exponents so I tried equal bases: x - 1 = 1/x and that allowed me to equate the exponents giving x^2 = x+1 which is same bloody equation. Both of which give your observation - the golden ratio. Pure blind luck. Sorry I can't be more of a help. This equation was obviously fudged.
Upon more thinking I realized that you can easily make up your own such equations, just making sure that equating the bases gives the same solution as equating the exponents.
1) (3x + 2)(2x -1) = 0 and this leads to
6x + 1 = 2/x (use these as the bases)
2) (5x+3)(2x-1) = 0 and this leads to
10x^2 = -x + 3 (use these as the exponenets)
3 ) Finally:
[6x+1]^(10x^2) = [2/x]^(-x+3)
which has solution x = .5 as it can be easily checked that LS = RS = 4^(5/2) = 32
Equations of this form have to satisfy (base1)^exponent1 = (base2)^exponent2 where base1 and base2 have no common powers. In our original equation base1 and base2 have a common power in 1/x. This forces the re-arrangement.
So now if you solve base1 = base2 (base1 and base2 having no common powers) and get a solution for x, that value of x also has to satisfy
exponent1 = exponent 2. If exponent1 <> exponent2 then there is no solution.
I stumbled across the answer. I re-arranged the original equation to:
[x-1]^(x^2/2) = [1/x]^(x + 1 - x^2/2)
Looked at that for a long time going nowhere.
Then I started thinking = bases and = exponents so I tried equal bases: x - 1 = 1/x and that allowed me to equate the exponents giving x^2 = x+1 which is same bloody equation. Both of which give your observation - the golden ratio. Pure blind luck. Sorry I can't be more of a help. This equation was obviously fudged.
Wow this is really cool!!
The answer is 𝜑.
I found it graphically. It's easy to verify, but I can't find the algebrical way from the initial equation.
Upon more thinking I realized that you can easily make up your own such equations, just making sure that equating the bases gives the same solution as equating the exponents.
1) (3x + 2)(2x -1) = 0 and this leads to
6x + 1 = 2/x (use these as the bases)
2) (5x+3)(2x-1) = 0 and this leads to
10x^2 = -x + 3 (use these as the exponenets)
3 ) Finally:
[6x+1]^(10x^2) = [2/x]^(-x+3)
which has solution x = .5 as it can be easily checked that LS = RS = 4^(5/2) = 32
Thanks for the problem.
An addendum:
Equations of this form have to satisfy (base1)^exponent1 = (base2)^exponent2 where base1 and base2 have no common powers. In our original equation base1 and base2 have a common power in 1/x. This forces the re-arrangement.
So now if you solve base1 = base2 (base1 and base2 having no common powers) and get a solution for x, that value of x also has to satisfy
exponent1 = exponent 2. If exponent1 <> exponent2 then there is no solution.