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* .

$(x-2{)}^{{\mathrm{log}}^{2}(x-2)+\mathrm{log}(x-2{)}^{5}-12}={10}^{2}\mathrm{log}(x-2)$

I tried solving by first considering $\mathrm{log}(x-2)$ as a variable, say $t$. Then I expressed $(x-2)$ as ${10}^{t}$ . Then after using some properties of log, I reached till here-

${10}^{{t}^{3}+5{t}^{2}-12t}={10}^{2}t$

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 $x=3,102,2+{10}^{-7}$

Recently I was going through a problem from the book Problems in Mathematics - *V Govorov & P Dybov* .

$(x-2{)}^{{\mathrm{log}}^{2}(x-2)+\mathrm{log}(x-2{)}^{5}-12}={10}^{2}\mathrm{log}(x-2)$

I tried solving by first considering $\mathrm{log}(x-2)$ as a variable, say $t$. Then I expressed $(x-2)$ as ${10}^{t}$ . Then after using some properties of log, I reached till here-

${10}^{{t}^{3}+5{t}^{2}-12t}={10}^{2}t$

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

Beginner2022-07-22Added 17 answers

Equations such as

$f(t)={10}^{{t}^{3}+5{t}^{2}-12t-2}-t=0$

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

${y}_{1}(t)={10}^{{t}^{3}+5{t}^{2}-12t-2}$

${y}_{2}(t)=t$

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

${y}_{1}(t)=\frac{1}{100}-\frac{3}{25}t\mathrm{log}(10)+O\left({t}^{2}\right)$

which has a negative slope while ${y}_{2}(t)$ has a positive slope. Using this expansion gives another estimate close to

$t=\frac{1}{4(25+3\mathrm{log}(10))}\simeq 0.00783509$

So, let us define the overall function

$f(t)={10}^{{t}^{3}+5{t}^{2}-12t-2}-t$

and let us try to find its roots starting from a given estimate ${t}_{0}$. Newton procedure will update this guess accodring to

${t}_{n+1}={t}_{n}-\frac{f({t}_{n})}{{f}^{\prime}({t}_{n})}$

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.

$f(t)={10}^{{t}^{3}+5{t}^{2}-12t-2}-t=0$

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

${y}_{1}(t)={10}^{{t}^{3}+5{t}^{2}-12t-2}$

${y}_{2}(t)=t$

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

${y}_{1}(t)=\frac{1}{100}-\frac{3}{25}t\mathrm{log}(10)+O\left({t}^{2}\right)$

which has a negative slope while ${y}_{2}(t)$ has a positive slope. Using this expansion gives another estimate close to

$t=\frac{1}{4(25+3\mathrm{log}(10))}\simeq 0.00783509$

So, let us define the overall function

$f(t)={10}^{{t}^{3}+5{t}^{2}-12t-2}-t$

and let us try to find its roots starting from a given estimate ${t}_{0}$. Newton procedure will update this guess accodring to

${t}_{n+1}={t}_{n}-\frac{f({t}_{n})}{{f}^{\prime}({t}_{n})}$

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.

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