Finding intersection between functions containing logarithms 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.
Thanks!
 A: 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 $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
A: 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:
$x=8\log(x)$
But the "closed form" is probably not what you expected as it involves the lambert w function:
Solution:
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:
$$-\frac{\ln 10}{8}=u \ln u=e^{\ln u}\ln u$$
Therefore:
$$\ln u=W(-\frac{\ln 10}{8})$$
$$u=\frac{1}{x}=e^{W(-\frac{\ln 10}{8})}$$
$$x=e^{-W(-\frac{\ln 10}{8})}$$
Now,
$$W(u)e^{W(u)}:=u$$
Thus,
$$\frac{W(u)}{u}=e^{-W(u)}$$
And we can rewrite our solutions as:
$$x=-\frac{W(-\frac{\ln 10}{8})}{\frac{\ln 10}{8}}$$
A: 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.
