Tower of $x$

Algebra Level 2

$\large x^{x^{x^{x^{.^{.^{.}}}}}} = 2$

What is the value of $x?$ Put your answer up to 2 decimal places.

The answer is 1.41.

2 solutions

Gabriel Quintero
Dec 7, 2017

Using logarithm we get:

$x^{x^{x^{x^{x...}}}} = 2$

$log_x x^{x^{x^{x^{x...}}}} = log_x 2$

$x^{x^{x^{x^{x...}}}}log_x x = log_x 2$

Since $x^{x^{x^{x^{x...}}}} = 2$

We get:

$2 log_x x = log_x 2$

$2 = log_x 2$

$x^{2} = 2$

$x = \sqrt{2}$

By your logic, $x^{x^{x^{x^{x...}}}} = 4$ gives a solution of $x = \sqrt2$ as well. So if $x= \sqrt2$ , then what is the value of $x^{x^{x^{x^{x...}}}}$ ? 2 or 4?

Pi Han Goh - 3 years, 6 months ago

First of all, I'm really excited you replied, this is my second solution in Briliant, so pardon any mistakes please. I'm using the information given so it would be 2.

Gabriel Quintero - 3 years, 6 months ago

A Former Brilliant Member
Dec 1, 2017

We have $x^{x^{x^{x^{x...}}}} = 2$ . Now, to what power do you raise $x$ ? To $x^{x^{x^{x^{x...}}}}$

We already set that equal to 2, and when you plug it in, you get: $x^{2} = 2$ or $x=\sqrt{2}$

By your logic, $x^{x^{x^{x^{x...}}}} = 4$ gives a solution of $x = \sqrt2$ as well. So if $x= \sqrt2$ , then what is the value of $x^{x^{x^{x^{x...}}}}$ ? 2 or 4?

Pi Han Goh - 3 years, 6 months ago

$x^{x^{x^{x... } } } = 2$ $x^{2} = 2$ $x^{4} = 4$

The answer is $\sqrt{2}$ , you just squared both sides.

A Former Brilliant Member - 3 years, 5 months ago

You didn't answer my question. I asked what is the possible value(s) of $x^{x^{x^{\ldots}}}$ . I didn't ask for the solution for $x$ .

Pi Han Goh - 3 years, 5 months ago

Tower of x x x

The answer is 1.41.

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Tower of $x$