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Difference between revisions of "1950 AHSME Problems/Problem 42"

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==Solution==
 
==Solution==
  
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==Solution 1:==
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Taking the log, we get <math>\log_x 2 = x^{x^{x^{.^{.^.}}}}=2 </math>, and <math>\log_x 2 = 2</math>. Solving for x, we get <math>2=x^2</math>, and <math>\sqrt{2}=x \Rightarrow \mathrm{(D)}</math>
 
Taking the log, we get <math>\log_x 2 = x^{x^{x^{.^{.^.}}}}=2 </math>, and <math>\log_x 2 = 2</math>. Solving for x, we get <math>2=x^2</math>, and <math>\sqrt{2}=x \Rightarrow \mathrm{(D)}</math>
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==Solution 2:==
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<math>x^{x^{x^{.^{.^.}}}}=2</math> is the original equation. If we let <math>y=x^{x^{x^{.^{.^.}}}}</math>, then the equation can be written as <math>y=2</math>. This also means that <math>x^y=2</math>, considering that adding one <math>x</math> to the start and then taking that <math>x</math> to the power of <math>y</math> does not have an effect on the equation, since <math>y</math> is infinitely long in terms of <math>x</math> raised to itself forever. It is already known that <math>y=2</math> from what we first started with, so this shows that <math>x^y=x^2=2</math>. If <math>x^2=2</math>, then that means that <math>\sqrt{2}=x \Rightarrow \mathrm{(D)}</math>.
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This is just a faster way once you get used to it, instead of taking a log of the function.
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~mathmagical
  
 
==See Also==
 
==See Also==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 00:21, 14 January 2023

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

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 41 		Followed by
Problem 43
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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