Finding x according to a series

Question

Suppose $x > 0$. Define a sequence for $n\in\mathbb{Z^+}$ in the following way: $$a_1 = x\\ a_2 = x^x \\ a_3 = x^{x^x}\\ a_4 = x^{x^{x^x}}$$

If you know that $$\lim_{n\to\infty}a_n=5$$ then what is $x$?

Answer 1

$$a_{n+1}=x^{a_n}$$ Applying limit, $$\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}x^{a_n}$$ $$\lim_{n\to\infty}a_{n+1}=x^{\lim_{n\to\infty}a_n}$$ $$5=x^5$$ $$x=\sqrt[5]5$$

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Alternatively, observe that $$\lim_{n\to\infty}a_{n}=x^{x^{x^{x^{x^{x^{\ldots}}}}}}=x^{\lim_{n\to\infty}a_{n}}$$ then similarly, $$x^5=5$$

Answer 2

Write it out as

$$ x^{x^{x^{x^\ldots}}} = 5 $$ and notice that the exponent on $x$ is the same as the expression on the LHS. So we can rewrite as

$$ x^5 = 5 $$

and this is easily solved from here.