Playing with Graphs 6

I Did it my old way

Let x<0

e^(9x) = 3x^28

consider the function f(x) = e^9x / 3x^28

differentiate it

You will obtain that function increasing in the interval (-28/9,0) and decreases in (-infinity , -28/9)

Put x = -28/9 in above function y= e^9x/3x^28

you will get y<1 hence y=1 will cut the graph at two points

Similarly we can do for positive part.

But as @neelesh vij pointed out as it is an even function number of solutions on left and right hand side of x axis will be same hence making a total of 4 solutions

$f(x) = |e^{|9x|} - 3x^{28}|$

As function is symmetric about Y- axis so number of solutions in $x>0$ will be equal to no. of solutions in $x<0$

So, Let $x>0$

$\Rightarrow 9x = 28 \ln (3x)$

$\Rightarrow \dfrac{9}{28} = \dfrac{\ln (3x)}{x}$

Now differentiating the function $\dfrac{\ln (3x)}{x}$ to get its critical points:

$\dfrac{d}{dx} \left( \dfrac{\ln x}{x} \right) = \dfrac{1 - \ln (3x)}{x^2}$ which is positive for $0<x<\dfrac e3$ so it has its maximum value at $x= \dfrac e3$

Max value $= \dfrac 3e > \dfrac {9}{28}$

Now the function $\dfrac{\ln (3x)}{x}$ approaches $-\infty$ when $x \to 0$ and to $0$ when $x \to \infty$

So, clearly it has 2 solutions in the region $x>0$

So, total no. of solutions = $\boxed{4}$