Range of solutions to the log equation

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Range of solutions to the log equation

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obihs

#1 h Apr 1, 2025, 6:18 PM

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Let $n$ be a positive integer, and consider the equation:
$(\log x)^n - x + 1 = 0\quad\cdots(\heartsuit)$ Answer the following questions. You may assume that $2.7<e<2.72$ is known.

$(1)\quad$ Determine the number of real solutions of equation $(\heartsuit)$ for each $n$ .

$(2)\quad$ For $n\ge 3$ , let $r_n$ be the segond largest real solution of $(\heartsuit)$ .

$(\i)\quad$ Find $\alpha$ such that $\lim_{n\to\infty} r_n =\alpha.$

$(\i\i)\quad$ Find $\lfloor\beta\rfloor$ , where $\beta$ is defined as

$\lim_{n\to\infty}n(r_n-\alpha)=\beta.$

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#2 h Apr 3, 2025, 12:31 AM • 1 Y

Y by obihs

Note that $\heartsuit'=\frac{n\ln^{n-1}x}{x}-1$ and $\heartsuit''=\frac{n(n-1)\ln^{n-2}x-nx\ln^{n-1}x}{x^2}=\frac{n(n-1-x\ln x)\ln^{n-2}x}{x^2}$

$\heartsuit''=0\implies x\ln x=n-1$ . Since $(x\ln x)'=\ln x+1$ is monotonic increasing with $\heartsuit''$ has minimum $-\frac{1}{e}$ at $x=\frac{1}{e}$ . Note that $\lim_{x\to 0^+}x\ln x=0$ and $\lim_{x\to\infty}x\ln x=\infty$ . So $\heartsuit''=0$ has only one solution $x^*$ .

This means $\heartsuit'$ is monotonic on $(0,x^*)$ and monotonic on $(x^*,\infty)$ , so $\heartsuit$ has at most one critical point in each of these intervals.

But also note $\lim_{x\to\infty}\ln^n x=\infty$ and $\lim_{x\to 0^+}\ln^nx=\begin{cases}\infty&\quad\text{even }n\\-\infty&\quad\text{odd }n\end{cases}$ . Also $\heartsuit'(1)=-1$ and $\heartsuit(1)=0$ .

Therefore at $n\geq 3$ when $n\geq 3$ , $\heartsuit=0$ has $2$ real solutions for $n$ even, and $3$ real solutions for $n$ odd.

At $n=1$ and $n=2$ only one root.

The second largest solution is $r_n=1$ , so $\alpha=\lim_{n\to\infty} r_n=1$ and $\alpha=\lim_{n\to\infty} n(r_n-\alpha)=0$ .

It is more interesting for the largest root of $\heartsuit$ , since that solution is decreasing with limit $e$ , and the smallest solution in the odd $n$ cases, since that solution is decreasing with limit $\frac{1}{e}$ .

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obihs

#3 h Apr 4, 2025, 7:44 AM

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@rchokler
Thank you for answering!!!!
But actually, when $n=3$ , there are $4$ real solutions of $(\heartsuit)$ .
about $x=0.438, 1, 4.688, 96.258$
I hope this helps you

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#4 h Apr 4, 2025, 9:19 AM • 1 Y

Y by obihs

Nice problem! Let $f(x) = \ln(x)^n - x + 1$ . We have $f'(x) = n \ln(x)^{n-1} \frac 1 x - 1$ and $f''(x) = n \ln(x)^{n-2} \frac 1 {x^2} (n-1 - \ln(x))$ . So $f$ is concave on $[e,e^{n-1}]$ and convex on $[e^{n-1},\infty)$ . On the other hand $f(e) = 2-e < 0$ , $f(e^{n-1}) = (n-1)^n - e^{n-1} + 1 > 0$ for $n \ge 4$ and $f(x) \to -\infty$ for $x \to \infty$ . By convexity/concavity this implies that we have exactly one zero in $[e,e^{n-1}]$ and exactly one zero in $[e^{n-1},\infty)$ .

For the asymptotics of the second largest zero $r_n$ we note that $f(e) < 0$ and
$f(e^{1+\frac e n}) = (1 + \frac e n)^n - e^{1 + \frac e n} + 1 \to e^e -e+1 > 0 \text{ for } n \to \infty.$ This implies that $r_n \in [e,e^{1 + \frac e n}]$ for $n$ sufficiently large. This already gives $\alpha = e$ . For finer asymptotics we write $r_n = e^{1 + \frac {s_n} n}$ . By the above $s_n \in [0,e]$ and in particular $s_n$ is bounded. We have
$e^{s_n} \sim (1 + \frac {s_n} n)^n = e^{1 + \frac {s_n} n} - 1 \to e-1 \text{ for } n \to \infty,$ which implies that $s_n \to \ln(e-1)$ for $n \to \infty$ and thus
$n(r_n - e) = n( e^{1 + \frac {s_n} n} - e) = ne ( e^{\frac {s_n} n} - 1) \to e \ln(e-1).$

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