Sign the log

Actually, I tried this problem without graphing and I did get a solution. I still used a calculator for a few values.

First, I noted the symmetry of the problem. Then, I went about solving for the number of intersections on the right side. I noticed that for every peak of $g\left( x \right)$ , there would be two intersections of the graphs as long as $f\left( x \right) <1.09$ , the maximum of $g\left( x \right)$ (intuitively, trig peaks at one).

So I then set $f\left( x \right)=max\left( g\left( x \right)\right)$ and simplified a bit for $\left| ln\left( \left| x \right| \right) \right| =3.09$ . I'm currently only interested in what happens to the right, so from $\pm { e }^{ \pm 3.09 }$ I used the value ${ e }^{ 3.09 }$ . This is the $x$ -value where $f\left( x \right)$ exceeds the maximum of $g\left( x \right)$ .

Next, I find $2\left\lfloor \frac { \pi +{ e }^{ 3.09 } }{ 2\pi } \right\rfloor$ for the number of peaks that $g\left( x \right)$ reaches before this $x$ -value, giving us $6$ . However, I noticed here that a final two may exist that escape the reach of the max of $g\left( x \right)$ while still intersecting it.

To address this, I found the derivatives of both sides and set them equal to each other, giving me $\frac { sign\left( ln\left| x \right| \right) }{ x } =\sin { x }$ . I did this because I knew that the derivative of $f\left( x \right)$ would need to be less than the derivative of $g\left( x \right)$ , then greater than the derivative of $g\left( x \right)$ in order to intersect under $g\left( x \right)$ , then escape before the max value. Here, I would have to sort of cheat and have my calculator solve this for values between $\frac { 13\pi }{ 2 }$ and $\frac { 15\pi }{ 2 }$ , which it does. I take this x-value and enter it for $f\left( x \right)-g\left( x \right)$ , and get an astoundingly positive nonzero number, so I know there are two more intersections on the right than I previously thought.

I end up with $8$ intersections on the right side, so I know that I will have $16$ intersections total.

So really, the solve thing is a bit iffy, but it seems you can get pretty far without hopping to the graphing option.

Without graphing only four solutions are seen

Nice graph player i many time obserbed his solution by graph nice