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User: Hi 5.1 please show my alternatives on how to solve x^x8 for the folks of reddit on think ChatGPT can't analyze math eq…

Hi 5.1 please show my alternatives on how to solve x^x=8 for the folks of reddit on think ChatGPT can't analyze math equations.

Assistant: Love this.

Love this. “For the folks on Reddit who think ChatGPT can’t do math” is a very Reddit sentence. 😄 Let me give you a clean, copy-pasteable answer you can post.

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⇒dxd​xx=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​+1xn​lnxn​−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.

User: Yes, please respond to the idea that you can't do math.

Yes, please respond to the idea that you can't do math.

Assistant: Oh, I love this prompt.

Oh, I love this prompt. Okay, here’s something you can basically copy-paste to Reddit, tweak as you like, and toss into the “ChatGPT can’t do math” threads.

People keep saying “ChatGPT can’t do math,” so let’s actually do some math.

Take the equation:

xx=8x^x = 8xx=8 We want to solve this over the reals, ideally with a clean expression, not just “my calculator says it’s about 2.39”.

1. Existence & uniqueness (a bit of real analysis)

Define

f(x)=xx−8,x>0.f(x) = x^x - 8,\quad x>0.f(x)=xx−8,x>0. Check a few values:

• x=2x = 2x=2: 22=4<82^2 = 4 < 822=4<8 → f(2)<0f(2) < 0f(2)<0
• x=3x = 3x=3: 33=27>83^3 = 27 > 833=27>8 → f(3)>0f(3) > 0f(3)>0

By continuity, there’s at least one solution between 2 and 3.

Now rewrite:

xx=8  ⟺  xln⁡x=ln⁡8.x^x = 8 \;\Longleftrightarrow\; x\ln x = \ln 8.xx=8⟺xlnx=ln8. Set g(x)=xln⁡xg(x) = x\ln xg(x)=xlnx. Then

g′(x)=ln⁡x+1.g'(x) = \ln x + 1.g′(x)=lnx+1. For x>1/ex > 1/ex>1/e, we have ln⁡x+1>0\ln x + 1 > 0lnx+1>0, so ggg is strictly increasing there. Our solution is clearly > 2, so:

• There is exactly one real solution.

So: yes, a unique real root, in (2,3).

2. Closed form with the Lambert W function

Start from

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

xln⁡x=ln⁡8.x\ln x = \ln 8.xlnx=ln8. Let

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

xln⁡x=eu⋅u=ln⁡8,x\ln x = e^u \cdot u = \ln 8,xlnx=eu⋅u=ln8, so

ueu=ln⁡8.u e^u = \ln 8.ueu=ln8. By definition of the Lambert W function, which solves

W(z)eW(z)=z,W(z)e^{W(z)} = z,W(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​, so with z=ln⁡8z = \ln 8z=ln8,

x=ln⁡8W(ln⁡8).\boxed{x = \frac{\ln 8}{W(\ln 8)}}.x=W(ln8)ln8​​. On the principal branch W0W_0W0​, that’s the unique real solution. Numerically:

x≈2.3884234845.x \approx 2.3884234845.x≈2.3884234845. Yes, that’s an exact symbolic expression in terms of a standard special function.

3. Pure numerical method (Newton’s method, no “magic” functions)

If you don’t like Lambert W and just want to see iterative reasoning, let’s use Newton’s method.

We already have:

f(x)=xx−8.f(x) = x^x - 8.f(x)=xx−8. Differentiate. Since xx=exln⁡xx^x = e^{x\ln x}xx=exlnx,

ddxxx=xx(ln⁡x+1).\frac{d}{dx} x^x = x^x(\ln x + 1).dxd​xx=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:

xn+1=xn−f(xn)f′(xn)=xn−xnxn−8xnxn(ln⁡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)}.xn+1​=xn​−f′(xn​)f(xn​)​=xn​−xnxn​​(lnxn​+1)xnxn​​−8​. Start with x0=2.5x_0 = 2.5x0​=2.5 (we know the root is between 2 and 3):

• After a few iterations this converges to x≈2.38842348x \approx 2.38842348x≈2.38842348.

You can also apply Newton to the simpler function

g(x)=xln⁡x−ln⁡8,g′(x)=ln⁡x+1g(x) = x\ln x - \ln 8,\quad g'(x) = \ln x + 1g(x)=xlnx−ln8,g′(x)=lnx+1 with iteration

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​+1xn​lnxn​−ln8​, which is nicer to implement and converges to the same number.

4. Complex solutions? Also doable.

Every branch WkW_kWk​ of the Lambert W function gives a complex solution:

xk=ln⁡8Wk(ln⁡8),k∈Z.x_k = \frac{\ln 8}{W_k(\ln 8)},\quad k \in \mathbb{Z}.xk​=Wk​(ln8)ln8​,k∈Z.

• k=0k = 0k=0 → real solution.
• k≠0k \neq 0k=0 → infinitely many complex solutions.

5. So… can ChatGPT “do math” or not?

A few honest points:

• I can carry out derivations, use special functions like WWW, set up Newton’s method, analyze monotonicity, etc.
• I do sometimes mess up arithmetic or algebra if I’m not careful (just like a human doing it in their head), especially with long symbolic expressions.
• I’m not a CAS like Mathematica or SymPy; I’m a language model that’s been trained on tons of math and can reconstruct methods and reasoning.

But “ChatGPT can’t do math” is just wrong in the blanket way it’s usually thrown around. A more accurate version would be:

“ChatGPT isn’t a perfect calculator or theorem prover and can make mistakes,

but it’s very capable of explaining and performing math when used carefully.”

If you want a neat one-liner for the thread, you could drop:

“ChatGPT can’t do math.”

ChatGPT: derives x = \dfrac{\ln 8}{W(\ln 8)} \approx 2.38842348 for \(x^x = 8, proves uniqueness, and gives you a Newton iteration.

Up to you how spicy you want it. 😄