Marisol Rivers

2022-07-21

Solution for this Logarithmic Equation
Recently I was going through a problem from the book Problems in Mathematics - *V Govorov & P Dybov* .
$\left(x-2{\right)}^{{\mathrm{log}}^{2}\left(x-2\right)+\mathrm{log}\left(x-2{\right)}^{5}-12}={10}^{2}\mathrm{log}\left(x-2\right)$
I tried solving by first considering $\mathrm{log}\left(x-2\right)$ as a variable, say $t$. Then I expressed $\left(x-2\right)$ as ${10}^{t}$ . Then after using some properties of log, I reached till here-
${10}^{{t}^{3}+5{t}^{2}-12t}={10}^{2}t$
or
${10}^{{t}^{3}+5{t}^{2}-12t-2}=t$
Now I have no idea how to approach further. The answer in the references says $x=3,102,2+{10}^{-7}$

Sandra Randall

Equations such as
$f\left(t\right)={10}^{{t}^{3}+5{t}^{2}-12t-2}-t=0$
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
${y}_{1}\left(t\right)={10}^{{t}^{3}+5{t}^{2}-12t-2}$
${y}_{2}\left(t\right)=t$
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
${y}_{1}\left(t\right)=\frac{1}{100}-\frac{3}{25}t\mathrm{log}\left(10\right)+O\left({t}^{2}\right)$
which has a negative slope while ${y}_{2}\left(t\right)$ has a positive slope. Using this expansion gives another estimate close to
$t=\frac{1}{4\left(25+3\mathrm{log}\left(10\right)\right)}\simeq 0.00783509$
So, let us define the overall function
$f\left(t\right)={10}^{{t}^{3}+5{t}^{2}-12t-2}-t$
and let us try to find its roots starting from a given estimate ${t}_{0}$. Newton procedure will update this guess accodring to
${t}_{n+1}={t}_{n}-\frac{f\left({t}_{n}\right)}{{f}^{\prime }\left({t}_{n}\right)}$
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.

Do you have a similar question?