Solving a polynomial with a natural log included with other terms for zero

Question

This problem has been plaguing me. Can the following equation be solved for zero? When I graph it, it clearly has solutions, but I'm struggling to solve on paper.

[(x-5)^2] - ln(x) = 0

Answer 1

Your equation is not solvable in terms of elementary functions or Lambert W. You need to solve your equation numerically.

You need the term closed-form.

$$(x-5)^2-\ln(x)=0$$ $$(x-5)^2=\ln(x).$$ We see, your equation cannot have a solution that is an algebraic number because for algebraic $x$, the left-hand side of the latter equation is an algebraic number whereas the right-hand side is a transcendental number.

$x\to e^t$, for all real $t$: $$(e^t)^2-10e^t-t+25=0$$ The function on the left-hand side of this equation is an algebraic function in dependence of both $t$ and $e^t$. Liouville proved that such kind of functions (over a complex domain without isolated points) don't have (partial) inverses that are elementary functions.

The equation is also an algebraic equation in dependence of both $t$ and $e^t$. Lin proved, assuming Schanuel's conjecture is true, that such kind of equations don't have solutions except $0$ that are elementary numbers.

$$\frac{1}{5+\sqrt{t}}\ e^t=1$$

We see, the equation is not solvable in terms of Lambert W and poosibly not in terms of generalized Lambert W.