Showing that $\ln(x)^2+2(x-1)\ln x-3x+1=0$ has only $2$ real solutions

Question

Is there any elementary way to show that $\ln (x)^2+2(x-1)\ln x-3x+1=0$ has $2$ real solutions on $(0,\infty)$?

I did it by this way.

Let $f(x)=(\ln x)^2+2(x-1)(\ln x)-3x+1$.

signs of $f(\frac{1}{2})$, $f(1)$, $f(e^2)$ are $+$, $-$, $+$ since $f(x)$ is continuous, due to this changes in sign there should be at least two roots for $f(x)=0$.

I want to know if there's any nicer way to do this.

Answer 1

As you have already noticed, you have shown that at least $2$ real solutions exist, to show that only 2 real solutions exist we can use that

$$(\ln(x))^2+2(x-1)(\ln x)-3x+1=0 $$

$$\iff \ln x=\frac{-2(x-1)\pm\sqrt{4(x-1)^2-4(-3x+1)}}{2}$$

$$\iff \ln x =(1-x)\pm\sqrt{x^2+x}$$

and then, by derivation, show that functions $g(x)=(1-x)+\sqrt{x^2+x}$ and $h(x)=(1-x)-\sqrt{x^2+x}\,$ for $x>0$ are monotonic, respectevely increasing and decreasing, and from here conclude for the existence of only $2$ real solutions, invoking IVT and using that $\log x$ is monotonic itself and surjective onto $\mathbb R$.

Answer 2

Consider the function$$f(x)=\log ^2(x)+2 (x-1)\, \log (x)-3 x+1$$for which $$f'(x)=-\frac{(x+2)-2 (x+1)\, \log (x)}{x}$$ Now, look at $$g(x)=(x+2)-2 (x+1)\, \log (x)$$ $$g'(x)=\frac{2 (x+1)}{x}+2 \log (x)-1$$

$$g'(x)=0 \qquad \implies \qquad x=-\frac{1}{W\left(-\sqrt{e}\right)}$$ where $W(.)$ is Lambert function which, in the real domain exists if $x\geq -\frac 1e$.

So $g'(x)$ is positive, $g(x)$ is an increasing function and $f'(x)$ is then also increasing.

Then, only one root for $f'(x)$

Answer 3

We can consider finding the roots of $ \ (\ln x)^2 + 2(x-1)\ln x - 3x + 1 \ = \ 0 \ $ as equivalent to determining the intersections of the curve for $ \ f(x) \ = \ (\ln x)^2 + 2(x-1)\ln x \ $ with the line $ \ y \ = \ 3x - 1 \ \ . \ $ The first two derivatives of the first function are $$ f'(x) \ \ = \ \ 2·\frac{\ln x}{x} \ + \ 2·\ln x \ - \ \frac{2}{x} \ + \ 2 \ \ = \ \ \left( \frac{2}{x} \ + \ 2 \right)·\ln x \ - \ \left( \frac{2}{x} \ - \ 2 \right) \ \ , $$ $$ f''(x) \ \ = \ \ - \frac{2}{x^2} ·\ln x \ + \ \frac{2}{x^2} \ + \ \frac{2}{x} \ + \ \frac{2}{x^2} \ \ = \ \ \frac{2·( \ 2 \ + \ x \ - \ \ln x \ )}{x^2} \ \ . $$

The domain of $ \ f(x) \ $ is $ \ x \ > \ 0 \ \ $ and it is the case that $ \ x + 2 \ > \ \ln x \ \ , \ $ so $ \ f''(x) \ > \ 0 \ $ always (or $ \ f(x) \ $ is convex or "concave upward"). "By inspection", we see that $ \ f'(x) \ = \ 0 \ $ for $ \ x \ = \ 1 \ \ , \ $ so $ \ f(x) \ $ has a single and absolute minimum at $ \ (1 \ , \ 0) \ \ . $

The line $ \ y \ = \ 3x - 1 \ $ has its $ \ x-$intercept at $ \ \left(\frac13 \ , \ 0 \right) \ \ , \ $ so we may expect it to intersect the curve for $ \ f(x) \ $ once in the interval $ \ \frac13 \ < \ x \ < \ 1 \ $ and once in the interval $ \ x \ > \ 1 \ \ . \ $

In fact, if we take $ \ \ln 2 \ \approx \ 0.7 \ \ , \ $ we can estimate that $$ \ f \left(\frac12 \right) \ \approx \ (-0.7)· \left[ \ (-0.7) + 2·\left(\frac12 - 1 \right) \ \right] \ = \ 1.2 \ \ , \ \ f' \left(\frac12 \right) \ \approx \ 6·(-0.7) - 2 \ = \ -6.2 \ \ ; $$ $$ f(4) \ \approx \ (1.4)· [ \ (1.4) + 2·(4 - 1) \ ] \ = \ 10.4 \ \ , \ \ f' (4 ) \ \approx \ \frac52·(1.4) + \frac32 \ = \ 5 \ \ . \ $$ Since $ \ 3·\frac23 \ - \ 1 \ = \ 1 \ \ , \ $ one root is in the interval $ \ \frac12 \ < \ x \ < \frac23 \ \ ; \ $ we can linearize $ \ f(x) \ $ to produce $$ 1.2 \ - \ 6.2·\left(x \ - \ \frac12 \right) \ \ = \ \ 1 \ + \ 3·\left(x \ - \ \frac23 \right) \ \ \Rightarrow \ \ x \ \ \approx \ \ 0.576 \ \ . $$ For the other root, with $ \ 3·4 - 1 \ = \ 11 \ \ , \ $ linearization about $ \ x \ = \ 4 \ $ yields $$ 10.4 \ + \ 5·(x \ - \ 4 ) \ \ = \ \ 11 \ + \ 3·(x \ - \ 4) \ \ \Rightarrow \ \ x \ \ \approx \ \ 4.3 \ \ . $$ [More precise values of the roots obtained from WolframAlpha are $ \ 0.5819 \ $ and $ \ 4.3678 \ \ . \ ] $