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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?

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    $\begingroup$ It seems unlikely that $x$ is allowed to be an arbitrary real value if you haven't learned about induction yet. If $x=\sqrt2$, what is your understanding of $(4\sqrt2)!$? Isn't $4x$ constrained to be a nonnegative integer? $\endgroup$
    – saulspatz
    Jul 30 at 15:20
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    $\begingroup$ This question can't be answered, in the manner of Wolfram Alpha, unless you know about the gamma function. It's not a pre-calculus question. If you restrict $4x$ to be a nonnegative integer, then to show there are no solutions, think about divisibility. What primes divide the left-hand side? What are the possible values of the right-hand side? $\endgroup$
    – saulspatz
    Jul 30 at 15:34
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    $\begingroup$ To 5 dp, the positive roots are empirically $0.13294$ and $2.15375$. It's not hard by considering the asymptotic behaviour (including sign, based on the parity of $x$) of $\Gamma(x+1)$ either side of negative integer $x$ that infinitely many negative roots exist, but they're bound to require numerical methods to compute. $\endgroup$
    – J.G.
    Jul 30 at 16:25
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    $\begingroup$ What kind of a solution are you after? If you're not allowing the Gamma function, then $x!$ can only really make sense for integers $x \geq 0$, so $(4x)!$ can only make sense for numbers of the form $n/4$ with $n \geq 0$ an integer. And it is quite easy to check that there are no solutions of this form. $\endgroup$
    – Joppy
    Sep 8 at 5:45
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    $\begingroup$ @JitendraSingh How do I take $(4 \times 0.13294)!$? What number is that, and how can it be evaluated using class 10 methods? $\endgroup$
    – Joppy
    Sep 8 at 5:49
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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?

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    $\begingroup$ Impressive answer. +1 $\endgroup$
    – Sebastiano
    Jul 30 at 16:42
  • $\begingroup$ The last term reminds me of the residue function. What a shame that the $n$-th power is also taken in the denominator. $\endgroup$
    – vitamin d
    Jul 31 at 1:15
  • $\begingroup$ @vitamind I just plugged into the Lagrange Inversion Theorem formula. In reality, the “denominator” always has $f^{-1}(x)-f^{-1}(x_0)$. The link has more context. What value of “a” could work here? $\endgroup$ Jul 31 at 1:31
  • $\begingroup$ Not the one I wanted but still best among others so here is the bounty $\endgroup$
    – user876009
    Sep 12 at 16:07
  • $\begingroup$ @JitendraSingh Thanks, you could split me the bounty and accept the other answer. $\endgroup$ Sep 12 at 16:14
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Taking in consideration that you haven't been taught induction yet you are searching for a natural value of $x \in (0,1,2,\dots)$ as factorial is defined for natural numbers.

Let's suppose that we have an x number that satisfy our equation, we have that $4^{4x}-4^x=4^{x}(4^{3x}-1)=(4x)!$ we can notice that $4^{3x}-1$ is odd as it is an even number minus an odd one, then the biggest power of $2$ that can divide $4^{x}(4^{3x}-1)$ is the same as the one that can divide $4^x$ . It is easy to see that it is $2^{2x}=4^x$. Now lets notice than for $(4x)!$ the biggest power of two that divide is much bigger and that is because it is at least $2x$, as $(4x)!=1\cdot 2 \cdot 3 \cdots (4x-1)\cdot 4x$ and we can notice that every $2$ numbers one is divisible by $2$, they will be $(2,4,6,8\cdot 4x-2, 4x)$ We can also notice that some of them are divisible for bigger powers of two. Here we have a contradiction because if $4^{x}(4^{3x}-1)=(4x)!$, every single number that divides one side of the equation has to divide the other one (because it is an equality) but we know that is not true for some power of two.

Then there is no natural number that satisfy the above equation.

note: if you want to find the exact power of two that divides $(4x)!$ there is something call legendre's formula that finds the exact value but it is not needed for this demonstration.

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    $\begingroup$ This seems to be the type of solution OP was looking for. $\endgroup$ Sep 11 at 3:29

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