1950 AHSME Problems/Problem 42

Problem

The equation $x^{x^{x^{.^{.^.}}}}=2$ is satisfied when $x$ is equal to:

$\textbf{(A)}\ \infty \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt[4]{2} \qquad \textbf{(D)}\ \sqrt{2} \qquad \textbf{(E)}\ \text{None of these}$

Solution

Solution 1:

Taking the log, we get $\log_x 2 = x^{x^{x^{.^{.^.}}}}=2$ , and $\log_x 2 = 2$ . Solving for x, we get $2=x^2$ , and $\sqrt{2}=x \Rightarrow \mathrm{(D)}$

Solution 2:

$x^{x^{x^{.^{.^.}}}}=2$ is the original equation. If we let $y=x^{x^{x^{.^{.^.}}}}$ , then the equation can be written as $y=2$ . This also means that $x^y=2$ , considering that adding one $x$ to the start and then taking that $x$ to the power of $y$ does not have an effect on the equation, since $y$ is infinitely long in terms of $x$ raised to itself forever. It is already known that $y=2$ from what we first started with, so this shows that $x^y=x^2=2$ . If $x^2=2$ , then that means that $\sqrt{2}=x \Rightarrow \mathrm{(D)}$ .

This is just a faster way once you get used to it, instead of taking a log of the function.

~mathmagical

Solution 3:

Given $x^{x^{x^{.^{.^.}}}}=2$ , we see that $x$ is also raised to $x^{x^{x^{.^{.^.}}}}$ , so $x^2 = 2$ $\Rightarrow x=\sqrt{2} \mathrm{{D}}$

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 41	Followed by Problem 43
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All AHSME Problems and Solutions

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