Staircase
Solving the equation x^3 = 3 is easy but what about solving equations with a 'staircase' of powers?
Age
16 to 18
Challenge level
Problem
This problem was suggested by Yatir Halevi.
Solve the equation:
$$x^{(x^3)}=3.$$
Solve the equation:
$$x^{(x^{(x^3)})}=3.$$
Can you find solutions to all equations of the form:
$$x^{x^{x^{x^{x^{x^{...^{a}}}}}}}=a$$
where the sequence of powers is defined in the same way and $a$ is a positive integer?
Explain what you have done and prove that you have found all possible solutions.
Getting Started
Consider this as a sequence given by:
$$x_{n+1}=x^{x_n}$$
where $x_1=x^3$.
Now consider the sequence of equations given by $x_n=3$.
Student Solutions
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 $x=\pm n^{1/n}$.
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 $x=\pm n^{1/n}$.