# Solution for this Logarithmic Equation

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

• I don't think this is solvable in terms of elementary functions. I've tried solving it numerically and got two possible answers: $3.0186$ and $83.2060$. – Tunococ Aug 3 '14 at 2:39
• $x=3$ is not a solution, the equation becomes $1=0$. – Ian Aug 3 '14 at 2:40
• Is $\log^2$ the square of the log, or the doubly-iterated log? That might explain it. (Maybe.) – Semiclassical Aug 3 '14 at 2:46
• That'd make the $x=3$ root correct, then. I similarly wonder if they meant $\log((x-2)^5)$ or $(\log(x-2))^5$. AND: Is that the log base 10 or base-$e$? – Semiclassical Aug 3 '14 at 2:53
• @Semiclassical they meant $\log((x−2)^5)$ or by property of log it can be written as $5\log(x-2)$ – Dysfunctional Aug 3 '14 at 2:56

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.

• Note that, as per the comments above, the equation has $\log^2(x)=\log(\log x)$. Because of that, the simplified LHS cited in the question isn't correct. – Semiclassical Aug 3 '14 at 4:13
• @Semiclassical. Obviously, you are correct. I did not see the subtility (I am almost blind). What do you think I do : delete my answer or just let it since I startted from OP's equation ? I would appreciate your opinion. Cheers :-) – Claude Leibovici Aug 3 '14 at 4:16
• Not sure. Your answer does make clear how this question is by far the exception rather than the rule as far as solvability, which is indeed useful. So myself I'd be fine with it remaining, either as a regular answer or as a CW. – Semiclassical Aug 3 '14 at 4:20
• No problem at all ! I prefer you don't know the number of mistakes I can do ! I let my answer as it is because, if the equation was correct, it would have been nice ! Cheers :-) – Claude Leibovici Aug 3 '14 at 5:48
• These are the correct answers as per the teacher I had consulted just an hour before. The answer given in the references are wrong and 3 is not a solution (it has to be assumed as square log, since no information has been given on it). Sorry for the inconvenience, as you had to give time to an already incorrect question and more than that, on incorrect information too.... you may close this question or downvote it, if you want. – Dysfunctional Aug 3 '14 at 5:51

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...).

• Thanks for your answer. But just now I have confirmed from a teacher that since it is not clear whether it is squared or iterated log, we must assume it to be squared only. And if we do that the equation is NOT solvable by high school methods. The answers provided are also wrong. But I would surely upvote your answer for the time you took for solving an incorrect question. – Dysfunctional Aug 3 '14 at 5:41