What is the intersection point of functions $f(x) = 8x^2$; $g(x) = 64x \log x$?

Question

I tried to solve this equation by hand without success. I used wolframalpha and the solution in the plot is (26.09, 5446.96). I just want to solve this equation step by step. Thanks in advance, cheers :DPlot wolframalpha

Answer 1

You have to compute the real numbers $x$ such that $f(x)=g(x)$, and then the intersection points are given by pairs $(x,f(x))$ for such real numbers $x$. The equation to solve is:

$8x^{2}=64x\log{x}$

The logarithm is not defined for $x=0$, so we may disregard that value of $x$ and obtain the new equation (also dividing by $8$):

$x=8\log{x}$

Now taking exponentials:

$e^{x}=x^{8}$

An equation of this type has no solution by hand: it is a transcendental equation. So the best think you can do, in my opinion, is to use some numerical method to compute the solution. For example, asking Wolfram Alpha as you already did.

Answer 2

Expanding on the answer from @Pedro, $$8x^2 = 64x\log x$$ is undefined if $x=0$, so $$\frac x8 = \log x$$ $$e^\frac x8 = x$$ $$-\frac18 = -\frac18 x e^{-\frac18 x}$$ Now we can use Lambert's W function, also called the Product Log, which is the inverse of $xe^x$ : $$W\left(-\frac18\right) = -\frac18 x$$ $$x = -8 W\left(-\frac18\right)$$ The W function has several branches (like the square root, positive or negative). From WolframAlpha, the two solutions are $$x_1 = -8*\text{ProductLog}[-1, -1/8] \approx 26.09$$ $$x_2 = -8*\text{ProductLog}[0, -1/8] \approx 1.15$$

Answer 3

If you do not want to use Lambert function, consider that you look for the zeros of $$f(x)=x-8\log(x)$$ $$f'(x)=1-\frac 8x$$ $$f''(x)=\frac 8{x^2} >0 \qquad \forall x$$

The first derivative cancels for $x=8$ and $f(8)=8-8\log(8)\approx -8.63553 <0$. So, there are two roots, the first lower than $8$, the second larger than $8$.

By inspection $f(1)=1$ and this could be a good starting point for Newton method (without any overshoot of the solution - by Darboux theorem). Then, for the first root, the iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.000000000 \\ 1 & 1.142857143 \\ 2 & 1.155291477 \\ 3 & 1.155370822 \\ 4 & 1.155370825 \end{array} \right)$$ For the second root, let us start using $x_0=16$ and get as iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 16.00000000 \\ 1 & 28.36141956 \\ 2 & 26.13113396 \\ 3 & 26.09349745 \\ 4 & 26.09348548 \end{array} \right)$$ Notice that could could have a better starting point using Taylor series around the minimum $$f(x)=(8-8 \log (8))+\frac{1}{16} (x-8)^2+O\left((x-8)^3\right)$$ which will give as approximation $$x_0=8 \left(1+\sqrt{2 (\log (8)-1)}\right)\approx 19.7545$$ but will not seriously affect the path to the solution.