At what point does $8x^2$ and $64x\log(x)$ intersect?

I'm trying to catch up on my math, but this stumped me. Something awry in my understanding of logarithms.

I figured that I could equate both functions and solve for $0$:

$$8x^2 = 64x\log(x)$$

then divide both sides by $8x$ for:

$$x = 8\log(x)$$

and then find:

$$10^x - x^8 = 0$$

Where did I go wrong? I seem to be stuck doing this by hand.


  • $\begingroup$ You cannot do it by hand (in the sense of finding roots of a polynomial), using standard functions. From your tags, you should probably be using a numerical method $\endgroup$ – David Peterson Jul 26 '16 at 0:25
  • $\begingroup$ Solving the intersection of a polynomial curve and a transcendental curve will only rarely have an explicit solution. One thing to try here might be Lambert's W function. $\endgroup$ – hardmath Jul 26 '16 at 0:25
  • $\begingroup$ Thanks guys. I was under the impression that if I could graph it and find that both functions intersect, then i should have been able to come up with an explicit solution by hand. Looking at Lambert's W function right now. $\endgroup$ – gswamibu Jul 26 '16 at 0:30
  • $\begingroup$ What do you mean where did you go wrong? Why do you think you did anything wrong? $\endgroup$ – fleablood Jul 26 '16 at 4:46

You didn't do anything wrong, except I would be careful of dividing by $8x$ for general algebra problems because what if $x=0$. But $\log (x)$ is undefined at $0$ so we are good.

Anyways you correctly reach:


But the "closed form" is probably not what you expected as it involves the lambert w function:


First convert from logarithms base $10$ to base $e$:

$$x=\frac{8 \ln x}{\ln 10}$$

Then the substitution $x=\frac{1}{u}$ gives:

$$\frac{1}{u}=-\frac{8}{\ln 10} \ln u$$

It can be rearranged to this:

$$-\frac{\ln 10}{8}=u \ln u=e^{\ln u}\ln u$$


$$\ln u=W(-\frac{\ln 10}{8})$$

$$u=\frac{1}{x}=e^{W(-\frac{\ln 10}{8})}$$

$$x=e^{-W(-\frac{\ln 10}{8})}$$





And we can rewrite our solutions as:

$$x=-\frac{W(-\frac{\ln 10}{8})}{\frac{\ln 10}{8}}$$

  • 1
    $\begingroup$ Wow, huge thank you for spending time on this. Looks like I'll be practicing lambert's function tonight until it becomes intuitive. I love how you detailed this step by step for us to refer back to this. $\endgroup$ – gswamibu Jul 26 '16 at 0:54
  • $\begingroup$ No problem, I just hope there's no careless mistakes. $\endgroup$ – Ahmed S. Attaalla Jul 26 '16 at 1:14

In fact, there two roots for the equation $x=8\log_{10}(x)$.

Consider the function $$f(x)=x-8\log_{10}(x)\implies f'(x)=1-\frac{8}{x \log (10)}\implies f''(x)=\frac{8}{x^2 \log (10)}$$ The first derivative cancels for $x=\frac{8}{\log (10)}$; it is negative if $x<\frac{8}{\log (10)}$, positive if $x>\frac{8}{\log (10)}$. The second derivative test shows that $x=8$ corresponds to a minimum. Since $f(8)=8-8\log_{10}(8)<0$, then there exactly two roots.

As Ahmed S. Attaalla answered, the analytical solutions are given in terms of Lambert function. They are $$x_1=-\frac{8 W\left(-\frac{\log (10)}{8}\right)}{\log (10)}\approx 1.57231$$ $$x_2=-\frac{8 W_{-1}\left(-\frac{\log (10)}{8}\right)}{\log (10)}\approx 6.50710$$ Sooner or later, you will learn about Lambert function and understand that any equation which can write $A+Bx+C\log(D+Ex)=0$ has analytical expressions in terms of it.

If you cannot use Lambert function, numerical methods would be required. Probably the simplest would be Newton method which, starting with a "reasonable" guess $x_0$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$


Draw the graphs of $y=x^8$ and $y=10^x$. It then becomes glaringly obvious that they intersect somewhere between $x=0$ and $x=-1$, BUT the equation you started with requires $x$ to be positive, since you're taking its logarithm. Do $y=x^8$ and $y=10^x$ intersect at some point where $x$ is positive? That's a slightly subtler question. However, look at what happens when $x$ goes from $1$ to $2$. You see that $x^8$ goes from $1$ to $256$ while $10^x$ goes from $10$ to $100$. Thus $x^8$ goes from less than $10^x$ to more than $10^x$. Therefore there is at least one intersection somewhere between $x=1$ and $x=2$. Just how many such intersections there are, and whether there are others not between $1$ and $2$, is a more involved question, which I might approach by using derivatives.

Probably the solution to your equation cannot be found algebraically and you'll need numerical methods such as Newton's method.

  • $\begingroup$ You bring up good points. I was thinking about implementing the newton method in python, but it seemed it would rob me of intuitively finding an intersection by hand. $\endgroup$ – gswamibu Jul 26 '16 at 0:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.