How to Solve the Equation (x)^x^3=3?

[racer]

Hello there

I've seen this question and tried to solve it but until now for no avail., I couldn't solve it but I know that the value of X is cubic root of 3

http://www.mathyards.com/attach/upload2/wh_45798340.bmp

Thanks.

 

[leon1127]

i don't think you can solve that analytically, instad you need to use numerical method?

 

[racer]

i don't think you can solve that analytically, instad you need to use numerical method?

Someone solved it by Calculus.

 

[HallsofIvy]

Do you have a reference for that? It certainly can be solved by "inspection"- noting that
[tex]\sqrt{3}^{\sqrt{3}^3}= \sqrt{3}^3= 3[/itex].<br /> <br /> But I have no idea what you <b>mean</b> by "solved it by Calculus"![/tex]

 

[racer]

But I have no idea what you mean by "solved it by Calculus"!

I didn't understand how he solved it but I've seen Transcendental Functions, he derived something, I recall that there was Inverses and then he got e^1/3, if you still have no idea, tell me and I'll bring the solution from him.

 

[D H]

Do you have a reference for that? It certainly can be solved by "inspection"- noting that
[tex]\sqrt{3}^{\sqrt{3}^3}= \sqrt{3}^3= 3[/itex].[/tex]

[tex] Surely you mean<br /> $$\sqrt[3] 3^{\sqrt[3] 3^3} = \sqrt[3] 3^3 = 3$$[/tex]

 

[HallsofIvy]

Yes, sorry. Very careless of me.

 

[John Creighto]

Surely you mean
$$\sqrt[3] 3^{\sqrt[3] 3^3} = \sqrt[3] 3^3 = 3$$

Seems easy to generalize:

$$x^{x^n}=n$$
$$\sqrt[n] n^{\sqrt[n] n \ ^n} = \sqrt[n] n \ ^n = n$$

 

[D H]

Seems easy to generalize:

$$x^{x^n}=n$$

$$\sqrt[n] n^{\sqrt[n] n \ ^n} = \sqrt[n] n \ ^n = n$$

An even more interesting generalization is

$$\underbrace{\sqrt[n]^n^{\sqrt[n]^n^{\dotsm^{\sqrt[n]^n}}}}_{n+1} = n$$

 

[John Creighto]

An even more interesting generalization is

$$\underbrace{\sqrt[n]^n^{\sqrt[n]^n^{\dotsm^{\sqrt[n]^n}}}}_{n+1} = n$$

Why is it n+1 levels?

 

[John Creighto]

Not sure if this is interesting but changin the original problem somewhat:

$$x^{x^n}=m$$

And assuming a solution of the form:

$$x= \sqrt[n]{nq}$$

We find that q is determined as follows:

$$(nq)^q=m$$

 

[sutupidmath]

it doesn't seem to matter at all how many levels you take it. for example

$$\sqrt [5]^5^{(\sqrt [5]^5)}^{5}=5$$

 

[D H]

it doesn't seem to matter at all how many levels you take it. for example

$$\sqrt [5]^5^{(\sqrt [5]^5)}^{5}=5$$

That is just a specific instance of John's generalization:

$$\sqrt[n] n^{\sqrt[n] n \ ^n} = \sqrt[n] n \ ^n = n$$

With $n$ as the final power it does not matter how many levels you take it. Start with the tautology:

$$\sqrt[n] n \ ^n = n$$

Substituting $\sqrt[n] n ^n$ for any of the ns on the left hand side of the tautology will yield a valid expression. Substituting for the power yields

$$\sqrt[n] n^{\sqrt[n] n \ ^n}= n$$

This substitution can be carried on indefinitely:

$$\sqrt[n] n ^{\sqrt[n] n ^{\cdots^{\sqrt[n] n \ ^n}}}= n$$

My "more interesting" generalization (post #9) replaces the final power ($n$) with with $\sqrt[n] n$. Changing the final power from $n$ to $\sqrt[n] n$ like I did in post #9 yields expressions like

$$\sqrt 2^{\sqrt 2^{\sqrt 2}} = 2$$

and

$$\sqrt[3] 3^{\sqrt[3] 3^{\sqrt[3]3^{\sqrt[3]3}}} = 3$$

 

[John Creighto]

My "more interesting" generalization (post #9) replaces the final power ($n$) with with $\sqrt[n] n$. Changing the final power from $n$ to $\sqrt[n] n$ like I did in post #9 yields expressions like

$$\sqrt 2^{\sqrt 2^{\sqrt 2}} = 2$$

and

$$\sqrt[3] 3^{\sqrt[3] 3^{\sqrt[3]3^{\sqrt[3]3}}} = 3$$

How do you show that?

 

[sutupidmath]

$$\sqrt[3] 3^{\sqrt[3] 3^{\sqrt[3]3^{\sqrt[3]3}}} = 3$$

Don't you think this should read like this:

$$\sqrt[3] 3^{\sqrt[3] 3^{\sqrt[3]3^{\sqrt[3]3}}} = \sqrt [3]^3$$

 

[D H]

How do you show that?

Apologies first: I should have used parentheses. Without parentheses, exponentiation evaluates right-to-left. Post #9 is correctly written as

$$\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1} = n$$

Taking the log (base n) of the left hand side,

$$\log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}\Biggr) =<br /> \sqrt[n] n \,<br /> \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_n\Biggr) =<br /> {\sqrt[n] n}^{\;2} \,<br /> \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n-1}\Biggr) =<br /> \cdots =<br /> {\sqrt[n] n}^{\;n} \log_n \left(\sqrt[n] n\right) = n\cdot \frac 1 n = 1$$

 

[sutupidmath]

Apologies first: I should have used parentheses. Without parentheses, exponentiation evaluates right-to-left. Post #9 is correctly written as

$$\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1} = n$$

Taking the log (base n) of the left hand side,

$$\log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}\Biggr) =<br /> \sqrt[n] n \,<br /> \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_n\Biggr) =<br /> {\sqrt[n] n}^{\;2} \,<br /> \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n-1}\Biggr) =<br /> \cdots =<br /> {\sqrt[n] n}^{\;n} \log_n \left(\sqrt[n] n\right) = n\cdot \frac 1 n = 1$$

WOW, nice trick!

 

[John Creighto]

Apologies first: I should have used parentheses. Without parentheses, exponentiation evaluates right-to-left. Post #9 is correctly written as

$$\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1} = n$$

Taking the log (base n) of the left hand side,

$$\log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}\Biggr) =<br /> \sqrt[n] n \,<br /> \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_n\Biggr) =<br /> {\sqrt[n] n}^{\;2} \,<br /> \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n-1}\Biggr) =<br /> \cdots =<br /> {\sqrt[n] n}^{\;n} \log_n \left(\sqrt[n] n\right) = n\cdot \frac 1 n = 1$$

Given where you placed the parenthesis, I still don’t follow.

 

[D H]

Given where you placed the parenthesis, I still don’t follow.

Step by step, then. The problem is to evaluate

$$\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}$$

To do so I will take the base-n log of this expression. Using $\log a^b = b\log a$

$$\log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}\Biggr) =<br /> \sqrt[n] n \,<br /> \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_n\Biggr)$$

We still have something of the form $\log a^v$. Applying the identity again yields

$${\sqrt[n] n}^{\;2} \,<br /> \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n-1}\Biggr)$$

After doing this $m<n$ times, the expression becomes

$${\sqrt[n] n}^{\;m} \,<br /> \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1-m}\Biggr)$$

After applying this identity n times, the log (base n) of the original expression becomes

$${\sqrt[n] n}^{\;n} \log_n \left(\sqrt[n] n\right)$$

The first term, ${\sqrt[n] n}^{\;n}$ is $n$. The second term, $\log_n \left(\sqrt[n] n\right)$ is $1/n$. The product of these two terms is one. If the log (base n) of the expression in question is one, the expression itself must be equal to n.

 

[racer]

But I have no idea what you mean by "solved it by Calculus"!

I meant that it was solved using Calculus, I got the solution and I uploaded it.

http://florble.co.uk/files/26341.JPG
http://florble.co.uk/files/8312.JPG
http://florble.co.uk/files/26783.JPG

 

[Vid]

He solved for a critical point. (aka when y'(x) = 0.) Not y(x) = 0.

 

[D H]

I meant that it was solved using Calculus, I got the solution and I uploaded it.

One issue with the original question is how to interpret the expression $x^{x^3}$. Exponentiation is not associative. Regardless of where one puts the parentheses in this expression, the uploaded solution does not solve the equation

$$x^{x^3}= 3$$

It finds where the function

$$f(x) = x^{(x^3)}$$

reaches a minimum. This function does reach a minimum at $$x=\exp(-1/3)$$. However, the value of the function at this point is approximately 0.8848, not 3.

The original problem has different solutions depending on interpretation, neither of which is $\exp(-1/3)$.

$$x^{(x^3)} = 3 \quad\Rightarrow \quad x \approx 1.44225$$

$$\left(x^x\right)^3 = 3 \quad\Rightarrow \quad x = \sqrt[3]3$$

 

[CRGreathouse]

The original problem has different solutions depending on interpretation

I thin k that x^x^3 is ambiguous. I don't think that $$x^{x^3}$$ is ambiguous at all; it's $$x^{(x^3)}$$.

$$x^{(x^3)} = 3 \quad\Rightarrow \quad x \approx 1.44225$$

$$\left(x^x\right)^3 = 3 \quad\Rightarrow \quad x = \sqrt[3]3$$

$$x^{(x^3)} = 3$$ is solved by $$x=\sqrt[3]3=1.44224957\ldots$$

$$(x^x)^3 = 3$$ is solved by $$x=1.319788817878158965558893060240296\ldots$$

 

[D H]

I thin k that x^x^3 is ambiguous. I don't think that $$x^{x^3}$$ is ambiguous at all; it's $$x^{(x^3)}$$.

Yes. Without parentheses, exponentiation is evaluated right to left (or top to bottom). Rationale: $$(a^b)^c = a^{(b*c)}$$

$$(x^x)^3 = 3$$ is solved by $$x=1.319788817878158965558893060240296\ldots$$

I solved both equations ($$x^{(x^3)}=3$$ and $$(x^x)^3 = 3$$) but pasted the same number in twice.

 

[CRGreathouse]

I solved both equations ($$x^{(x^3)}=3$$ and $$(x^x)^3 = 3$$) but pasted the same number in twice.

I thought that was the case (since the decimal expansion you posted works for the cube root of 3) but I wanted to clarify for others reading the thread.

 

[D H]

CR, thanks for the correction.

One can of course "use calculus" to solve either of the equations $$x^{x^3} = x^{\left(x^3\right)} = 3$$ or $$\left(x^x\right)^3 = x^{3x} = 3$$. It's called Newton's method. In general, to solve an expression of the form $$f(x)-a=0$$, Newton's method starts with an estimate of the solution and refines these estimates via

$$x_{n+1} = x_n + \frac{a-f(x_n)}{f'(x_n)}$$

The former expression, $$x^{x^3} = 3$$, has an algebraic solution: $$x=\sqrt[3]3$$. A numerical estimation technique will yield 1.44224957..., but why bother? The latter expression, $$\left(x^x\right)^3 = 3$$, is (most likely) transcendental; solving it numerically is (most likely) the only way to determine a value.

 

[uart]

My "more interesting" generalization (post #9) replaces the final power ($n$) with with $\sqrt[n] n$. Changing the final power from $n$ to $\sqrt[n] n$ like I did in post #9 yields expressions like

$$\sqrt 2^{\sqrt 2^{\sqrt 2}} = 2$$

and

$$\sqrt[3] 3^{\sqrt[3] 3^{\sqrt[3]3^{\sqrt[3]3}}} = 3$$

How do you show that?

I think he meant to write :

$$\sqrt 2^{\sqrt 2^{\sqrt {2}^2}} = 2$$

and

$$\sqrt[3] 3^{\sqrt[3]3^{\sqrt[3]3^{\sqrt[3]{3}^3}}} = 3$$

 

[D H]

No. I meant to write

$$\left (\sqrt 2 ^{\sqrt 2}\right)^{\sqrt 2} = 2$$

$$\left( \left(\sqrt[3]3 ^{\sqrt[3]3}\right)^{\sqrt[3]3}\right)^{\sqrt[3]3} = 3$$

I left out the parentheses, which are absolutely essential as exponentiation is non-associative.

 

[uart]

Ok I see. Yes that is more interesting.

 

[John Creighto]

Step by step, then. The problem is to evaluate

$$\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}$$

To do so I will take the base-n log of this expression. Using $\log a^b = b\log a$

$$\log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}\Biggr) =<br /> \sqrt[n] n \,<br /> \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_n\Biggr)$$

Sound you get instead:

$$\log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}\Biggr) =<br /> \underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_n\Biggr \,<br /> \log_n\left(\sqrt[n] n \right)$$