I'm trying to solve the equation $\log^2 (x^2) + \log (x-1) = 0$ but all I could do is to show that $1 < x < 2$. Wolfram Alpha says that $x = 1.508554…$, this is good, but I really want to write $x$ with some explicit expression, not numerical approximation.
This problem came from a group of friends, no one knows how to solve it.
Thanks!
Best Answer
There is no known closed-form solution for equations like this one. Let's simplify it: $$\log^2x^2+\log(x-1)=0$$ $$\log x^2\cdot\log x^2=-\log(x-1)$$ $$\log x^{2\log x^2}=\log\dfrac1{x-1}$$ $$x^{4\log x}=\dfrac1{x-1}$$ $$x^{-4\log x}=x-1$$ We have exponential and logarithmic function on LHS and linear function on RHS. When we have linear and exponential or logarithmic function in same equation we must use Lambert W function. Only equation which can be solved using Lambert W function is $$a^{f(x)}+bf(x)+c=0$$ if we can find $f^{-1}(x)$ where $a,b,c$ are constants. Equation $\log^2x^2+\log(x-1)=0$ cannot be reduced to this form, so there are no closed-form solution.