Editing Openai/69481a9b-0c20-800f-9469-35a475c239b3 (section)

==== We want to solve ====

xx=8x^x = 8xx=8
over the reals (and optionally say something about complex solutions).

===== Define =====

f(x)=xx−8,x>0.f(x) = x^x - 8,\quad x>0.f(x)=xx−8,x>0.
Some quick values:
* x=1x = 1x=1: 11=1<81^1 = 1 < 811=1<8 → f(1)=−7f(1) = -7f(1)=−7
* x=2x = 2x=2: 22=4<82^2 = 4 < 822=4<8 → f(2)=−4f(2) = -4f(2)=−4
* x=3x = 3x=3: 33=27>83^3 = 27 > 833=27>8 → f(3)=19f(3) = 19f(3)=19

Since fff is continuous on (0,∞)(0,\infty)(0,∞), by the Intermediate Value Theorem there is at least one root in (2,3)(2,3)(2,3).

For x>0x > 0x>0,

g(x)=xln⁡xg(x) = x\ln xg(x)=xlnx
is strictly increasing for x>1/ex > 1/ex>1/e (you can check g′(x)=ln⁡x+1>0g'(x) = \ln x + 1 > 0g′(x)=lnx+1>0 for x>1/ex > 1/ex>1/e). That means xln⁡x=ln⁡8x\ln x = \ln 8xlnx=ln8 has at most one solution for x>1/ex>1/ex>1/e. Since our root is between 2 and 3, the real solution is unique.

So: exactly one real solution, somewhere between 2 and 3.

===== Start from =====

xx=8,x>0.x^x = 8,\quad x>0.xx=8,x>0.
Take natural logs:

xln⁡x=ln⁡8.x\ln x = \ln 8.xlnx=ln8.
Now set

u=ln⁡x⇒x=eu.u = \ln x \quad\Rightarrow\quad x = e^u.u=lnx⇒x=eu.
Plug in:

xln⁡x=eu⋅u=ln⁡8.x\ln x = e^u \cdot u = \ln 8.xlnx=eu⋅u=ln8.
So we get

ueu=ln⁡8.u e^u = \ln 8.ueu=ln8.
By definition of the Lambert W function (which solves W(z)eW(z)=zW(z)e^{W(z)} = zW(z)eW(z)=z), we have:

u=W(ln⁡8).u = W(\ln 8).u=W(ln8).
Recall u=ln⁡xu = \ln xu=lnx, so

ln⁡x=W(ln⁡8)⇒x=eW(ln⁡8).\ln x = W(\ln 8)
\quad\Rightarrow\quad
x = e^{W(\ln 8)}.lnx=W(ln8)⇒x=eW(ln8).
There’s a standard identity:

eW(z)=zW(z).e^{W(z)} = \frac{z}{W(z)}.eW(z)=W(z)z.
Using that with z=ln⁡8z = \ln 8z=ln8, we get

x=ln⁡8W(ln⁡8).x = \frac{\ln 8}{W(\ln 8)}.x=W(ln8)ln8.
On the principal branch W0W_0W0, this gives the unique real solution:

x=ln⁡8W0(ln⁡8)≈2.3884234845.x = \frac{\ln 8}{W_0(\ln 8)} \approx 2.3884234845.x=W0(ln8)ln8≈2.3884234845.
So the tidy “closed form” is:

x=ln⁡8W(ln⁡8).\boxed{x = \dfrac{\ln 8}{W(\ln 8)}}.x=W(ln8)ln8.

===== If you don’t want Lambert W and just want a numeric approximation, you can use Newton’s method directly. =====

Take

f(x)=xx−8.f(x) = x^x - 8.f(x)=xx−8.
We need its derivative. For x>0x>0x>0,

xx=exln⁡x⇒ddxxx=xx(ln⁡x+1).x^x = e^{x\ln x} \Rightarrow \frac{d}{dx} x^x = x^x(\ln x + 1).xx=exlnx⇒dxdxx=xx(lnx+1).
So

f′(x)=xx(ln⁡x+1).f'(x) = x^x(\ln x + 1).f′(x)=xx(lnx+1).
Newton’s iteration is:

xn+1=xn−f(xn)f′(xn)=xn−xnxn−8xnxn(ln⁡xn+1)=xn−1−8/xnxnln⁡xn+1.x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
= x_n - \frac{x_n^{x_n} - 8}{x_n^{x_n}(\ln x_n + 1)}
= x_n - \frac{1 - 8/x_n^{x_n}}{\ln x_n + 1}.xn+1=xn−f′(xn)f(xn)=xn−xnxn(lnxn+1)xnxn−8=xn−lnxn+11−8/xnxn.
Pick a starting guess in (2,3). Say x0=2.5x_0 = 2.5x0=2.5.

Do a few iterations (rounded to 12 decimals):
* x1≈2.400611678683x_1 \approx 2.400611678683x1≈2.400611678683
* x2≈2.388577900139x_2 \approx 2.388577900139x2≈2.388577900139
* x3≈2.388423509467x_3 \approx 2.388423509467x3≈2.388423509467
* x4≈2.388423484499x_4 \approx 2.388423484499x4≈2.388423484499

Already at n=4n=4n=4, we’re stable to many decimal places:

x≈2.388423484499.x \approx 2.388423484499.x≈2.388423484499.
You can also simplify things a bit by applying Newton to

g(x)=xln⁡x−ln⁡8g(x) = x\ln x - \ln 8g(x)=xlnx−ln8
instead. Then

g′(x)=ln⁡x+1g'(x) = \ln x + 1g′(x)=lnx+1
and Newton becomes

xn+1=xn−xnln⁡xn−ln⁡8ln⁡xn+1.x_{n+1} 
= x_n - \frac{x_n \ln x_n - \ln 8}{\ln x_n + 1}.xn+1=xn−lnxn+1xnlnxn−ln8.
Same limit, easier derivative.

===== Another way to see what’s going on is to rewrite the equation as: =====

xx=8⟺xln⁡x=ln⁡8.x^x = 8 \quad\Longleftrightarrow\quad x\ln x = \ln 8.xx=8⟺xlnx=ln8.
So you can:
* Plot y=xln⁡xy = x\ln xy=xlnx.
* Plot the horizontal line y=ln⁡8y = \ln 8y=ln8.

Where they intersect is the solution. Since xln⁡xx\ln xxlnx is strictly increasing for x>1/ex > 1/ex>1/e, there is exactly one intersection point for x>0x > 0x>0, the same x≈2.388423x \approx 2.388423x≈2.388423.

Equivalently, you can plot y=xxy = x^xy=xx and the horizontal line y=8y=8y=8; again you’ll see a single intersection between 2 and 3.

===== If you care about complex solutions, each branch WkW_kWk of the Lambert W function gives you one: =====

xk=exp⁡(Wk(ln⁡8))=ln⁡8Wk(ln⁡8),k∈Z,x_k = \exp\bigl(W_k(\ln 8)\bigr) = \frac{\ln 8}{W_k(\ln 8)}, \quad k \in \mathbb{Z},xk=exp(Wk(ln8))=Wk(ln8)ln8,k∈Z,
so there are infinitely many complex solutions, one per branch. Only the principal branch k=0k=0k=0 is real.

===== - Real solution (closed form): x=ln⁡8W(ln⁡8)≈2.3884234845.x = \dfrac{\ln 8}{W(\ln 8)} \approx 2.3884234845.x=W(ln8)ln8≈2.3884234845. =====
* Derived by turning xx=8x^x = 8xx=8 into xln⁡x=ln⁡8x\ln x = \ln 8xlnx=ln8, substituting u=ln⁡xu = \ln xu=lnx to get ueu=ln⁡8u e^u = \ln 8ueu=ln8, then using the Lambert W function.
* Numerically, Newton’s method converges quickly from any starting guess in (2,3)(2,3)(2,3).

If you want, I can also help you wrap this into a snappy Reddit comment with some “yes, the robot can do math” spice.