I think that your argument is quite correct. For sufficiently large x the term $-12 x^x$ will dominate the expression.

Limit[9^(3 x) - 12 x^x + 45 x^3 - 66, x -> Infinity]

is negative infinity. In fact, as you say the only terms that are consequential for this zero are the first two. We can get a first estimation of the position of the zero if we observe that $9^{3x}=9^3^x=729^x$. So the question is when this term is more or less as large as the $-12 x^x$. if we forget the -12 for a second that would happen at $x \approx 729$. This is of course a wee bit imprecise because of the 12, but we are relatively close to 726.

Let's make this a bit more precise. We only use the relevant terms and ask when does:

$9^{3x}=12 x^x$

as before

$$729^x=12 x^x$$ $$729^x=12^{1/x}^x x^x$$ $$729^x=(12^{1/x} x)^x$$

this is obviously true when

$$729 = (12^{1/x} x)$$

Which is an easy one for Mathematica:

sols = Solve[(9^3) == 12^(1/x) x, x]

We are interested in the first solution - the one close(r) to 729.

N[x /. sols[[1]], 20]

gives 726.51083890096132690, which is identical to the first digits of the "full" solution 726.51083890096132690 Mathematica obtains in one of my posts above. BTW, many more digits are correct. Certainly the first 2000.

Cheers,

Marco

This is all very interesting. I never had to set Max iterations or working precision that high before.

It would have been good for users if Wolfram Alpha had explained how it got the numbers. The code shown simply does not do the job -- it did not use FindRoot[] or discuss setting MaxIterations or Working Precision to very high values.

The value of the slope explains why plotting the function near the second solution yielded nothing.

I think that Jim's argument is quite correct. It looks as if we need a dramatically increased WorkingPrecision here:

sol = FindRoot[9^(3 x) - 12 x^x + 45 x^3 - 66 == 0, {x, 726.509, 726.513}, MaxIterations -> 1000000, WorkingPrecision -> 10000]

This is only the first part of the solution! With that we get:

9^(3 x) - 12 x^x + 45 x^3 - 66 /. sol
(*0.*10^-7917*)

which is a numerical zero. The function is just quite steep at that point.

N[D[9^(3 x) - 12 x^x + 45 x^3 - 66, x] /. sol, 10]
(*-6.325431641*10^2079*)

Best wishes, Marco

@George Woodrow III:

9^(3 x) - 12 x^x + 45 x^3 - 66 /. x->726 is positive and 9^(3 x) - 12 x^x + 45 x^3 - 66 /. x->727 is negative. Would there not need to be a value of zero somewhere between 726 and 727 ?

Jim

@max ron: Maybe you should consider the following:

f[x_] := 9^(3 x) - 12 x^x + 45 x^3 - 66
Sign[f[Rationalize[726.5105]]]
(* 1 *)
Sign[f[Rationalize[726.5110]]]
(* -1 *)

Between a positive value and a negative value there's got to be a zero.