Solving Exponent Problem: Find Value of x

[altamashghazi]

Homework Statement

x^X^x^x^x=2. find value of x.

Homework Equations

taking log both sides, but it makes a equation which i am not able to solve.

The Attempt at a Solution

x^(x)^4=2
x^4logx=log2. what next?

 

[Mentallic]

Homework Statement

x^X^x^x^x=2. find value of x.

Homework Equations

taking log both sides, but it makes a equation which i am not able to solve.

The Attempt at a Solution

x^(x)^4=2
x^4logx=log2. what next?

Is that

$$x^{x^{x^{x^x}}}$$

or

$$(((x^x)^x)^x)^x$$

?

 

[altamashghazi]

the fist one

 

[altamashghazi]

the first one

 

[Mentallic]

Was this a problem you just made up? In any case, I don't believe you'll be able to solve it using analytic functions.

If you take the log of both sides, it becomes

$$x^{x^{x^x}}\log x = \log 2$$

and not $x^4\log x$ as you suggested. Also, the power tower isn't

$$x^{x^4}$$ either. If you use Knuth's up arrow notation, then it would be equivalent to x^^5 where each ^ represents an up arrow.

 

[altamashghazi]

then what should i do how to solve

 

[Curious3141]

Funnily enough, you can solve the infinite power tower quite easily, provided a solution exists, e.g.

$$x^{x^{x^{x^{x}...}}} = 2 \implies x^2 = 2 \implies x = \sqrt{2}$$

But there's no way of solving a finite power tower equation algebraically. You can probably do it numerically, though, to get an approximate solution.

EDIT: you can use $x = \sqrt{2}$ as an excellent starting guess for a numerical solver for your finite power tower problem.

 

[Mentallic]

Use approximation methods, such as the
Bisection method: http://en.wikipedia.org/wiki/Bisection_method
Newton's method: http://en.wikipedia.org/wiki/Newton's_method[/PLAIN]

 

[sambarbarian]

Curious3141 , I am curious to know how x^x^x ... = 2 became x^2 = 2 .

 

[Mentallic]

If x^x^x... = 2 then if we add another x at the bottom of that tower power, we haven't changed anything since the power tower is infinitely high. Or in other words,
y^x^x^x... = 2 where y=x is still equivalent to x^x^x..., so we substitute 2 for x^x^x... since that's what we assumed it is equal to, and arrive at

y^2 = 2
y=x
x^2 = 2

Curious3141

I am curious...

Curious tends to instigate these feelings in all of us