Prove the equation $\ln(x) = \frac1 {x-1}$ has exactly 2 real solutions.
Hello all. I thought of defining the function $f(x)=x-e^{\frac 1 {x-1}}$ and showing it has only 2 single roots, though I am not sure on how to show it and I understand it is better to prove using Lagrange's theorem. Would be happy to get some help on that question, thanks in advance :)
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HINT
Consider and study the function $f(x)=(x-1)\log x-1$ using derivatives and IVT to prove that we have exactly 2 solutions for $f(x)=0$.
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Consider $f(x)=ln(x)- \frac{1}{1-x}$ and then differentiate it once and look at how many times this function changes monotony
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Start with this: On $(1,\infty)$, $x\mapsto \ln(x)$ is strictly increasing and increases without bound. On the same interval $x \mapsto 1/(x-1)$ decreases from $+\infty$ to $0$. Now apply the intermediate value theorem to see you have at least one root on $(1,\infty)$. The monotonicities can be used to see the root is unique. Now deal similarly on $(0,1)$.
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Without using a study function, you can look over $]0,1[$ and $]1,+\infty[$.
On both segments, $x\mapsto \frac{1}{x-1}$ and $x\mapsto \ln(x)$ are continuous respectively strictly decreasing and increasing : they can have on both segment at most one intersection point. Now looking on the values near the boundaries, and using a corollary of the Intermediate value theorem, you can conclude the existence of exactly one point of intersection on each segment, two in total.
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$\ln x$ does not exist for $x \le 0$ and $\frac 1{x-1}$ does not exist for $x = 1$.
So if there are solution they will exist on the intervals $(0,1)$ and $(1, \infty)$.
On these intervals the function $f(x) = \ln x - \frac 1{x-1}$ is continuous.
$f'(x) = \frac 1x - (-1)\frac 1{(x-1)^2} = \frac 1x + \frac 1{(x-1)^2} > 0$ so $f'(x)$ is increasing on the intervals $(0,1)$ and $(1,0)$. So if $f(x)$ ever "crosses the $x$-axis". It can only do so at most once in each interval.
Taking the limits of $f(x)$ as $x \to 0^+; x \to 1^-; x\to 1^+$ and $x\to \infty$ we will see that $f(x)$ will have negative and positive values on each of these intervals. So by the intermediate value theorem there will be values in these intervals where $f(x) = 0$.
And as $f'(x) > 0$ there can only at most one on each interval, so there will be exactly one on each interval.
And, obviously, if $f(x) = 0$ then $\ln x = \frac 1{x-1}$.
Consider the function $f$ defined on $(0,1)\cup(1,\infty)$ by $$f(x) = (x-1)\ln x\,.$$ We have $\lim_{x\to 0^+} f(x) = +\infty$, $\lim_{x\to 1^-} f(x)=\lim_{x\to 1^+} f(x) = 0$, and $\lim_{x\to +\infty} f(x) =+\infty$. Further, $f$ is differentiable twice on its domain, with $$ f''(x) = \frac{x+1}{x^2} > 0 $$ and $$ f'(x) = 1+\ln x - \frac{1}{x} $$ which has limit $0$ at $x=1$ (on both sides), $-\infty$ at $0^+$ and $+\infty$ at $\infty$.Combining the two (the limits of $f'$ and the fact that $f'$ is strictly increasing on $(0,1)$, and strictly increasing on $(1,\infty)$), $f' <0$ on $(0,1)$ and $f'>0$ on $(1,\infty)$, i.e. you have strict monotonicity of $f$ on both parts of its domain.
From this last thing, along with the limits of $f$, you get the result: $f$ takes the value $1$ exactly once on $(0,1)$, and exactly once on $(1,\infty)$.