# Solve (x)^x^3=3 find X

1. Mar 9, 2008

### 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

Thanks.

Last edited by a moderator: Apr 23, 2017
2. Mar 9, 2008

### leon1127

i dont think you can solve that analytically, instad you need to use numerical method?

3. Mar 9, 2008

### racer

Someone solved it by Calculus.

4. Mar 9, 2008

### HallsofIvy

Staff Emeritus
Do you have a reference for that? It certainly can be solved by "inspection"- noting that
$$\sqrt{3}^{\sqrt{3}^3}= \sqrt{3}^3= 3[/itex]. But I have no idea what you mean by "solved it by Calculus"! 5. Mar 9, 2008 ### racer I didn't understand how he solved it but I've seen Transcendental Functions, he derived something, I recall that there was Inverses and then he got e^1/3, if you still have no idea, tell me and I'll bring the solution from him. 6. Mar 9, 2008 ### D H Staff Emeritus Surely you mean [tex]\sqrt[3] 3^{\sqrt[3] 3^3} = \sqrt[3] 3^3 = 3$$

Last edited: Mar 9, 2008
7. Mar 9, 2008

### HallsofIvy

Staff Emeritus
Yes, sorry. Very careless of me.

8. Mar 9, 2008

### John Creighto

Seems easy to generalize:

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

Last edited: Mar 9, 2008
9. Mar 9, 2008

### D H

Staff Emeritus
An even more interesting generalization is

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

10. Mar 9, 2008

### John Creighto

Why is it n+1 levels?

11. Mar 9, 2008

### 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:

$$(nq)^q=m$$

12. Mar 9, 2008

### sutupidmath

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

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

13. Mar 10, 2008

### D H

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

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

Last edited: Mar 10, 2008
14. Mar 10, 2008

### John Creighto

How do you show that?

15. Mar 10, 2008

### sutupidmath

Don't you think this should read like this:

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

16. Mar 10, 2008

### D H

Staff Emeritus
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,

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

17. Mar 10, 2008

### sutupidmath

WOW, nice trick!!!!!

18. Mar 10, 2008

### John Creighto

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

19. Mar 10, 2008

### D H

Staff Emeritus
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 $\log a^b = b\log a$

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

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

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

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

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

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.

20. Mar 10, 2008

### racer

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

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

Last edited by a moderator: May 3, 2017