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
\sqrt{3}^{\sqrt{3}^3}= \sqrt{3}^3= 3[/itex].<br /> <br /> But I have no idea what you <b>mean</b> by &quot;solved it by Calculus&quot;!

 

[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
\sqrt{3}^{\sqrt{3}^3}= \sqrt{3}^3= 3[/itex].<br />

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

 

[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<br />

 

[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<br />

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<br />

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)<br />

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

<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)

After doing this m&lt;n times, the expression becomes

<br /> {\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)<br />

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

<br /> {\sqrt[n] n}^{\;n} \log_n \left(\sqrt[n] n\right)<br />

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&#039;(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)<br />

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)<br />

 

[John Creighto]

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.

No, you put them in but apparently in the wrong spot.

 

[sutupidmath]

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

probbably he meant to write it sth like this:

\log_n (...(((\sqrt[n]^n)^{\sqrt [n]^n})^{\sqrt [n]^n})^{\sqrt [n]^n}...)^{\sqrt [n]^n} =\sqrt[n]^n \log_n (...(((\sqrt[n]^n)^{\sqrt [n]^n})^{\sqrt [n]^n})^{\sqrt [n]^n}...)=\sqrt [n]^n^{2} \log_n(..((\sqrt[n]^n)^{\sqrt[n]^n})^{\sqrt[n]^n})..)..=...=\sqrt[n]^n^{n}log_n\sqrt[n]^n=n*\frac{1}{n}log_n n=1 and hence:



(...(((\sqrt[n]^n)^{\sqrt [n]^n})^{\sqrt [n]^n})^{\sqrt [n]^n}...)^{\sqrt [n]^n}=n

Does this make any sense??

 

[John Creighto]

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

probbably he meant to write it sth like this:

\log_n (...(((\sqrt[n]^n)^{\sqrt [n]^n})^{\sqrt [n]^n})^{\sqrt [n]^n}...)^{\sqrt [n]^n} =\sqrt[n]^n \log_n (...(((\sqrt[n]^n)^{\sqrt [n]^n})^{\sqrt [n]^n})^{\sqrt [n]^n}...)=\sqrt [n]^n^{2} \log_n(..((\sqrt[n]^n)^{\sqrt[n]^n})^{\sqrt[n]^n})..)..=...=\sqrt[n]^n^{n}log_n\sqrt[n]^n=n*\frac{1}{n}log_n n=1 and hence:



(...(((\sqrt[n]^n)^{\sqrt [n]^n})^{\sqrt [n]^n})^{\sqrt [n]^n}...)^{\sqrt [n]^n}=n

Does this make any sense??

You don't even need logs to show that sense

(x^a)^b=x^(ab)