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Telescoping Series

Find $S_r = 1^r + 2^r + 3^r + ... + n^r$ where r is any fixed positive integer in terms of $S_1, S_2, ... S_{r-1}$.

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Growing

Which is larger: (a) 1.000001^{1000000} or 2? (b) 100^{300} or 300! (i.e.factorial 300)

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Climbing Powers

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?

Staircase

Stage: 5 Challenge Level: Challenge Level:1
Ho Chung gives a solution here:

For the first two equations, the answer is the cube root of 3, by observation. You can simply substitute this value in the equation and verify that it is a solution. We have to show that these are the only possible solutions.

What follows is really an argument by contradiction. If we assume that there exists a solution less than the cube root of 3 we reach an impossible situation and likewise for one greater than the cube root of 3.

Define a sequence $(x_n)$ by $x_1=x^3$, $x_{n+1}=x^{x_n}$. Observe that for $x> 1$, if $x^3> 3$ then the sequence is strictly increasing, and if $x^3< 3$ then the sequence is strictly decreasing.

Now to solve $x^{(x^3)}=3$, we are imposing $x_2=3$, so the sequence becomes $x^3$, 3, $x^3, 3, ...$. Since we must have $x> 1$ and the sequence is neither strictly increasing nor strictly decreasing, we must have $x^3=3$. This also clearly works. So $x=\sqrt[3]{3}$.

Similarly, to solve $x^{(x^{(x^3)})}=3$, we have $x_3=3$, so the sequence becomes $x^3$,$x^{(x^3)}$, 3, $x^3, ... $ and so again we have $x=\sqrt[3]{3}$.

For the general equation $$x^{x^{x^{x^{x^{x^{...^{n}}}}}}}=n$$ where the sequence of powers is defined in the same way, and $n$ is a positive integer, we can use the same argument. The solution is the $n$-th root of $n$ if $n$ is odd and when $n$ is even there are two solutions $x=\pm n^{1/n}$.