Solution for this Logarithmic Equation

Question

Recently I was going through a problem from the book Problems in Mathematics - *V Govorov & P Dybov* . $$(x-2)^{\log^2(x-2)+\log(x-2)^5-12}=10^2\log(x-2)$$

I tried solving by first considering $\log(x-2)$ as a variable, say $t$. Then I expressed $(x-2)$ as $10^t$ . Then after using some properties of log, I reached till here- $$10^{t^3+5t^2-12t}=10^2t$$ or $$10^{t^3+5t^2-12t-2}=t$$ Now I have no idea how to approach further. The answer in the references says ${x=3, 102, 2+10^{-7}}$

Answer 1

Equations such as $$f(t)=10^{t^3+5t^2-12t-2}-t=0$$ cannot be solve using analytical methods and numerical methods, such as Newton, should be used.

As you probably notice, you are looking for the intersection of two curves, namely $$y_1(t)=10^{t^3+5t^2-12t-2}$$ $$y_2(t)=t$$

If you plot the functions on the same graph, you should notice a clear intersection around $t=1.9$. There is also a root close to $t=0$ since, around this value, a Taylor expansion gives $$y_1(t)= \frac{1}{100}-\frac{3}{25} t \log (10)+O\left(t^2\right)$$ which has a negative slope while $y_2(t)$ has a positive slope. Using this expansion gives another estimate close to $$t=\frac{1}{4 (25+3 \log (10))} \simeq 0.00783509$$

So, let us define the overall function $$f(t)=10^{t^3+5t^2-12t-2}-t$$ and let us try to find its roots starting from a given estimate $t_0$. Newton procedure will update this guess accodring to $$t_{n+1}=t_n-\frac{f(t_n)}{f'(t_n)}$$ For the first solution, let us start at $t_0=0$; Newton iterates are then : $0.00783509$, $0.00801852$, $0.0080186$ which is the solution for six significant figures.

For the second solution, let us start at $t_0=1.9$; Newton iterates are then : $1.91187$, $1.90970$, $1.90959$ which is again the solution for six significant figures.

Since, from your changes of variable $x=2+10^t$, the solutions are then $x=3.01864$ and $x=83.2064$ which are the values given by Tunococ.

Answer 2

Note: As noted in the comments above, the log used here is the common log (base-10) and the notation $\log^2$ is the twice-iterated log (not the squared log).

First, your route is correct in defining $t=\log(x-2)$. But the LHS after substitution is not quite right: Observe that $10^{\log^2(x-2)}=\log(x-2)=t$ not $(10^t)^{t^2}=10^{t^3}$. Hence the correct equation to solve is $$t\cdot 10^{5t^2-12t}=10^2 t$$ This gives an immediate $t=0\implies x=3$ solution. The remainder are found from the equation $5t^-12t=2$ to be $$t=\dfrac{1}{5}(6\pm\sqrt{46}\implies 10^{\frac{1}{5}(6\pm\sqrt{46})}-2\approx 358,-1.3$$

Note that this reproduces the $x=3$ root quoted but not the $$x=2+10^2,2+10^{-7}\implies t=2,-7$$ roots. This would follow, however, if we had the quadratic equation $t^2-5t-12=2$ instead (which is suspiciously similar to the equation we do get...).