Where Did I Go Wrong?

When approaching the equation x^{x^{x^{x^{x^{x^{x^{...}}}}}}} 4 with the goal of finding x, many solvers might infer that x sqrt{2}. However, this may lead to an apparent contradiction, which is a fascinating conundrum in mathematics. Let's dive deeper into this equation and the underlying principles.

Solving the Equation

First, let's denote the infinite power tower as y. Thus, we have:

y x^{y}

Given that y 4, substituting this into our equation, we get:

4 x^{4}

From this, we can solve for x by taking the fourth root of both sides:

x 4^{1/4}

Simplifying, we find:

4^{1/4} 2^{2/4} 2^{1/2} sqrt{2}

Therefore, it is correct to state that:

x sqrt{2}

However, substituting x sqrt{2} into the equation, we get:

sqrt{2}^{sqrt{2}^{sqrt{2}^{sqrt{2}^{sqrt{2}^{sqrt{2}^{...}}}}}} 2

Let's denote the power tower again as z.

z sqrt{2}^{z}

To solve for z, we take the logarithm:

log{z} z cdot log{sqrt{2}}

Since log{sqrt{2}} frac{1}{2} log{2}, we have:

log{z} frac{1}{2} z cdot log{2}

This is a transcendental equation that can be challenging to solve directly. However, if we check the assumption that:

z 2

we find:

2 sqrt{2}^{2} 2

This implies that:

z 2 is indeed a solution to the power tower with x sqrt{2}.

The Correct Interpretation

Thus, the original equation x^{x^{x^{x^{x^{x^{x^{...}}}}}}} 4 leads to x sqrt{2} and the value of the infinite power tower is 2 and not 4, if we start with x sqrt{2}.

It is important to note that:

x sqrt{2}^{sqrt{2}^{sqrt{2}^{...}}} 2 x sqrt{4} 2 In no case does 2 4 No contradiction exists here and the solution is valid within the context of the infinite power tower.

Presh Talwalker has a brilliant video on this matter, which further elucidates the intricacies of such equations.

Further Considerations

Let's consider another approach. For any complex number x a bi, the equation x^{x^{x^{...}}} 4 may have different solutions. However, when dealing with real numbers:

sqrt{a^2} |a|

Thus, for a 2, we have:

sqrt{2^2} sqrt{4} 2

And for a -2, we have:

sqrt{(-2)^2} sqrt{4} 2

This shows that:

x sqrt{4} is equivalent to x 2 The square root function returns the positive value by convention This does not lead to the contradiction 0 2 or 1 0

Therefore, the hypothesis that the equation has an answer is valid, and the solution is correct within the real number system.