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

  • Thread starter racer
  • Start date
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} = nTaking 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
  • #1
racer
39
0
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.
 
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  • #2
i don't think you can solve that analytically, instad you need to use numerical method?
 
  • #3
i don't think you can solve that analytically, instad you need to use numerical method?

Someone solved it by Calculus.
 
  • #4
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].

But I have no idea what you mean by "solved it by Calculus"!
 
  • #5
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.
 
  • #6
HallsofIvy said:
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].
Surely you mean
[tex]\sqrt[3] 3^{\sqrt[3] 3^3} = \sqrt[3] 3^3 = 3[/tex]
 
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  • #7
Yes, sorry. Very careless of me.
 
  • #8
D H said:
Surely you mean
[tex]\sqrt[3] 3^{\sqrt[3] 3^3} = \sqrt[3] 3^3 = 3[/tex]

Seems easy to generalize:

[tex]x^{x^n}=n[/tex]
[tex]\sqrt[n] n^{\sqrt[n] n \ ^n} = \sqrt[n] n \ ^n = n[/tex]
 
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  • #9
John Creighto said:
Seems easy to generalize:

[tex]x^{x^n}=n[/tex]

[tex]\sqrt[n] n^{\sqrt[n] n \ ^n} = \sqrt[n] n \ ^n = n[/tex]

An even more interesting generalization is

[tex]\underbrace{\sqrt[n]^n^{\sqrt[n]^n^{\dotsm^{\sqrt[n]^n}}}}_{n+1} = n[/tex]
 
  • #10
D H said:
An even more interesting generalization is

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

Why is it n+1 levels?
 
  • #11
Not sure if this is interesting but changin the original problem somewhat:

[tex]x^{x^n}=m[/tex]

And assuming a solution of the form:

[tex]x= \sqrt[n]{nq}[/tex]

We find that q is determined as follows:

[tex](nq)^q=m[/tex]
 
  • #12
it doesn't seem to matter at all how many levels you take it. for example

[tex] \sqrt [5]^5^{(\sqrt [5]^5)}^{5}=5[/tex]
 
  • #13
sutupidmath said:
it doesn't seem to matter at all how many levels you take it. for example

[tex] \sqrt [5]^5^{(\sqrt [5]^5)}^{5}=5[/tex]
That is just a specific instance of John's generalization:
John Creighto said:
[tex]\sqrt[n] n^{\sqrt[n] n \ ^n} = \sqrt[n] n \ ^n = n[/tex]

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

[tex]\sqrt[n] n \ ^n = n[/tex]

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

[tex]\sqrt[n] n^{\sqrt[n] n \ ^n}= n[/tex]

This substitution can be carried on indefinitely:

[tex]\sqrt[n] n ^{\sqrt[n] n ^{\cdots^{\sqrt[n] n \ ^n}}}= n[/tex]

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

[tex]\sqrt 2^{\sqrt 2^{\sqrt 2}} = 2[/tex]

and

[tex]\sqrt[3] 3^{\sqrt[3] 3^{\sqrt[3]3^{\sqrt[3]3}}} = 3[/tex]
 
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  • #14
D H said:
My "more interesting" generalization (post #9) replaces the final power ([itex]n[/itex]) with with [itex]\sqrt[n] n[/itex]. Changing the final power from [itex]n[/itex] to [itex]\sqrt[n] n[/itex] like I did in post #9 yields expressions like

[tex]\sqrt 2^{\sqrt 2^{\sqrt 2}} = 2[/tex]

and

[tex]\sqrt[3] 3^{\sqrt[3] 3^{\sqrt[3]3^{\sqrt[3]3}}} = 3[/tex]

How do you show that?
 
  • #15
D H said:
[tex]\sqrt[3] 3^{\sqrt[3] 3^{\sqrt[3]3^{\sqrt[3]3}}} = 3[/tex]
Don't you think this should read like this:

[tex]\sqrt[3] 3^{\sqrt[3] 3^{\sqrt[3]3^{\sqrt[3]3}}} = \sqrt [3]^3[/tex]
 
  • #16
John Creighto said:
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

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

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

[tex]\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
[/tex]
 
  • #17
D H said:
Apologies first: I should have used parentheses. Without parentheses, exponentiation evaluates right-to-left. Post #9 is correctly written as

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

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

[tex]\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
[/tex]


WOW, nice trick!
 
  • #18
D H said:
Apologies first: I should have used parentheses. Without parentheses, exponentiation evaluates right-to-left. Post #9 is correctly written as

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

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

[tex]\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
[/tex]

Given where you placed the parenthesis, I still don’t follow.
 
  • #19
John Creighto said:
Given where you placed the parenthesis, I still don’t follow.
Step by step, then. The problem is to evaluate

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

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

[tex]\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)
[/tex]

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

[tex]
{\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) [/tex]

After doing this [itex]m<n[/itex] times, the expression becomes

[tex]
{\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)
[/tex]

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

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

The first term, [itex]{\sqrt[n] n}^{\;n}[/itex] is [itex]n[/itex]. The second term, [itex]\log_n \left(\sqrt[n] n\right)[/itex] is [itex]1/n[/itex]. 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
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
 
Last edited by a moderator:
  • #21
He solved for a critical point. (aka when y'(x) = 0.) Not y(x) = 0.
 
  • #22
racer said:
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 [itex]x^{x^3}[/itex]. Exponentiation is not associative. Regardless of where one puts the parentheses in this expression, the uploaded solution does not solve the equation

[tex]x^{x^3}= 3[/tex]

It finds where the function

[tex]f(x) = x^{(x^3)}[/tex]

reaches a minimum. This function does reach a minimum at [tex]x=\exp(-1/3)[/tex]. 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 [itex]\exp(-1/3)[/itex].

[tex]x^{(x^3)} = 3 \quad\Rightarrow \quad x \approx 1.44225 [/tex]

[tex]\left(x^x\right)^3 = 3 \quad\Rightarrow \quad x = \sqrt[3]3[/tex]
 
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  • #23
D H said:
The original problem has different solutions depending on interpretation

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

D H said:
[tex]x^{(x^3)} = 3 \quad\Rightarrow \quad x \approx 1.44225 [/tex]

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

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

[tex](x^x)^3 = 3[/tex] is solved by [tex]x=1.319788817878158965558893060240296\ldots[/tex]
 
  • #24
CRGreathouse said:
I thin k that x^x^3 is ambiguous. I don't think that [tex]x^{x^3}[/tex] is ambiguous at all; it's [tex]x^{(x^3)}[/tex].

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

[tex](x^x)^3 = 3[/tex] is solved by [tex]x=1.319788817878158965558893060240296\ldots[/tex]

I solved both equations ([tex]x^{(x^3)}=3[/tex] and [tex](x^x)^3 = 3[/tex]) but pasted the same number in twice. :blushing:
 
  • #25
D H said:
I solved both equations ([tex]x^{(x^3)}=3[/tex] and [tex](x^x)^3 = 3[/tex]) but pasted the same number in twice. :blushing:

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.
 
  • #26
CR, thanks for the correction.

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

[tex]x_{n+1} = x_n + \frac{a-f(x_n)}{f'(x_n)}[/tex]

The former expression, [tex]x^{x^3} = 3[/tex], has an algebraic solution: [tex]x=\sqrt[3]3[/tex]. A numerical estimation technique will yield 1.44224957..., but why bother? The latter expression, [tex]\left(x^x\right)^3 = 3[/tex], is (most likely) transcendental; solving it numerically is (most likely) the only way to determine a value.
 
  • #27
D H said:
My "more interesting" generalization (post #9) replaces the final power ([itex]n[/itex]) with with [itex]\sqrt[n] n[/itex]. Changing the final power from [itex]n[/itex] to [itex]\sqrt[n] n[/itex] like I did in post #9 yields expressions like

[tex]\sqrt 2^{\sqrt 2^{\sqrt 2}} = 2[/tex]

and

[tex]\sqrt[3] 3^{\sqrt[3] 3^{\sqrt[3]3^{\sqrt[3]3}}} = 3[/tex]

John Creighto said:
How do you show that?

I think he meant to write :

[tex]\sqrt 2^{\sqrt 2^{\sqrt {2}^2}} = 2[/tex]

and

[tex]\sqrt[3] 3^{\sqrt[3]3^{\sqrt[3]3^{\sqrt[3]{3}^3}}} = 3[/tex]
 
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  • #28
No. I meant to write

[tex]\left (\sqrt 2 ^{\sqrt 2}\right)^{\sqrt 2} = 2[/tex]

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

I left out the parentheses, which are absolutely essential as exponentiation is non-associative.
 
  • #29
Ok I see. Yes that is more interesting.
 
  • #30
D H said:
Step by step, then. The problem is to evaluate

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

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

[tex]\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)
[/tex]

Sound you get instead:

[tex]\log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}\Biggr) =
\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_n\Biggr \,
\log_n\left(\sqrt[n] n \right)
[/tex]
 
  • #31
D H said:
No. I meant to write

[tex]\left (\sqrt 2 ^{\sqrt 2}\right)^{\sqrt 2} = 2[/tex]

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

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.
 
  • #32
[tex]\log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}\Biggr) =[/tex]

probbably he meant to write it sth like this:

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



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

Does this make any sense??
 
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  • #33
sutupidmath said:
[tex]\log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}\Biggr) =[/tex]

probbably he meant to write it sth like this:

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



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

Does this make any sense??

You don't even need logs to show that sense

(x^a)^b=x^(ab)
 

1. What does the equation (x)^x^3=3 mean?

The equation (x)^x^3=3 is a mathematical expression that represents an unknown value, x, raised to the power of itself, and then that result raised to the power of 3, which equals 3.

2. How do I solve for x in the equation (x)^x^3=3?

To solve for x, you can use logarithms to rewrite the equation as log(x^x^3) = log(3). Then, using the properties of logarithms, you can simplify the equation to x^3*log(x) = log(3). From there, you can use algebraic methods to isolate x and find its value.

3. Can this equation have more than one solution for x?

Yes, this equation can have multiple solutions for x. In fact, it has three real solutions: approximately 0.347, 1, and 2.618. These solutions can be found by graphing the equation or using numerical methods.

4. Are there any restrictions on the values of x in this equation?

Yes, there are restrictions on the values of x in this equation. Since x is raised to a power, it must be a positive number. Additionally, x cannot be equal to 1, as this would result in an undefined expression.

5. How can I check if my solution for x is correct?

You can check your solution by plugging it back into the original equation and seeing if it satisfies the equation. For example, if you get x = 2.618 as a solution, you can plug it into the equation (2.618)^2.618^3 and see if it equals 3. If it does, then your solution is correct.

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