Solve for $x$: $$ 4^{4x}-4^x=(4x)! $$ (for all real values)
What is the biggest problem is that induction is not taught to me yet and hence can't be used.
Attempt $1$:
Assume $4^x=t$. So we get $t^4-t=24 (x!)$.
Letting $u$ denote $x!$
$$ t^4-t-24u=0 \implies u=\frac{t}{24}(t^3-1).$$
Attempt $2$:
Taking $4^x$ out as common
$$ 4^x(4^{3x}-1)=4x(4x-1)(4x-1)...(2)(1) $$
However this also yielded nothing.
Attempt $3$:
What I tried was did it by brute force-
Our equation is $4^{4x}-4^x-(4x)!=0$
Let $x$ be $-1$ and only taking the LHS
$$ 4^{-4}-4^{-1}-(4(-1))! = \text{ Infinity }$$
We can exclude all negative cases as the value can't be negative but negative factorial don't exist. So putting the value of $0$: $$ 4^0-4^0-0! \implies -1 $$
Using value of $1$:
$$ 4^4-4-4! \implies 228 $$
This implies the value will be between 1 and 2. However Wolfram Alpha result (which I will discuss at end) don't agree with me.
So how do I find the answer to this question without induction? Also why didn't Approach 3 worked for me? Also according to Wolfram Alpha over here how can I find such complex roots? Please help to solve this question?
Here is an answer where I will attempt to answer with a recursive solution and using the Lagrange Inversion Theorem. Here is graphical proof. Notice the graph is chaotic, but still gives the right answer and converges to $x=.132…$=solution very quickly after only 1 recursion.
$$\mathrm{4^{4x}-4^x=(4x)!\implies a_0=x=log_4\left(4^{4x}-(4x)!\right)=a_1\implies x=\lim_{n\to\infty}a_n=a_\infty,a_{n+1}= log_4\left(4^{4a_n{}}-(4{a_n})!\right)\implies x=.132…= log_4\left(4^{4{log_4\left(4^{4(…)}-(4(…))!\right)}}-(4{log_4\left(4^{4(…)}-(4(…))!\right)})!\right),a_2= log_4\left(4^{4{log_4\left(4^{4x}-(4x)!\right)}}-(4{log_4\left(4^{4x}-(4x)!\right)})!\right)}$$
There is not really a good way to find the nth derivative, so here are the results around a. Just plug in x=0 because this is the inverse of $4^{4x}-4^x-(4x)!=0$. Note that a=0 created an undefined expression:
$$\mathrm{x=a+\sum_{n=1}^\infty \frac{\left(4^{4x}-4^x-(4x)!-4^{4a}+4^a+(4a)!\right)^n}{n!}\lim_{x\to a}\frac{d^{n-1}}{dx^{n-1}}\left[\left(\frac{x-a}{4^{4x}-4^x-(4x)!-4^{4a}+4^a+(4a)!}\right)^n\right]= a+\sum_{n=1}^\infty \frac{\left(-1-4^{4a}+4^a+(4a)!\right)^n}{n!}\lim_{x\to a}\frac{d^{n-1}}{dx^{n-1}}\left[\left(\frac{x-a}{4^{4x}-4^x-(4x)!-4^{4a}+4^a+(4a)!}\right)^n\right]}$$
Root finding algorithms come to mind when solving, but they require derivatives, so maybe another recursive definition would work. Lets see what happens with $y=4^x$. We can get a “closed form” using the Elementary Root function or Fixed point operator. These are actual operators defined by Wolfram Research and so are known accepted operators which is not calculus. The k subscript denotes the kth fixed point or root. Let’s say the convention is that when $k=0,1$ then the following and this Graphical Visualization holds:
$$\mathrm{4^{4x}-4^x-(4x)!=0\mathop\implies^{x=log_4(y)} 4^{4 log_4(y)}-4^{log_4(y)}-(4 log_4(y))!=y^4-y-\left(4log_4(y)\right)!=0\implies y=y^4-\left(4\log_4(y)\right)!\implies y=Root\left[y^4-\left(4\log_4(y)\right)!-y\right]_k= \left[y^4-\left(4\log_4(y)\right)!-y\right]_k\implies y=FixedPoint\left[y^4-\left(4\log_4(y)\right)!\right]_k \implies y=\left(y\right)^4-\left(4\log_4(y)\right)!= \left(\left(y\right)^4-\left(4\log_4(y)\right)!\right)^4-\left(4\log_4(\left(y\right)^4-\left(4\log_4(y)\right)!)\right)!},… $$
Now we have an “almost polynomial equation” where the roots can be solved as $4^y=x$. Note the graph gets chaotic, but the infinite solutions still work from the recursion because the recursion approach vertical lines which are the solutions equated to the identity function.
I would do the same methods on both the original and substituted equation, but this would be exhaustive. Please correct me and give me feedback!
What about complex number solutions?