Difference between revisions of "1950 AHSME Problems/Problem 42"
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==Problem== | ==Problem== | ||
| − | The equation <math> x^{x^{x | + | The equation <math> x^{x^{x^{.^{.^.}}}}=2 </math> is satisfied when <math>x</math> is equal to: |
<math>\textbf{(A)}\ \infty \qquad | <math>\textbf{(A)}\ \infty \qquad | ||
| Line 8: | Line 8: | ||
\textbf{(D)}\ \sqrt{2} \qquad | \textbf{(D)}\ \sqrt{2} \qquad | ||
\textbf{(E)}\ \text{None of these}</math> | \textbf{(E)}\ \text{None of these}</math> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | ==Solution 1:== | ||
| + | |||
| + | 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> | ||
| + | |||
| + | |||
| + | ==Solution 2:== | ||
| + | |||
| + | <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>. | ||
| + | |||
| + | This is just a faster way once you get used to it, instead of taking a log of the function. | ||
| + | |||
| + | ~mathmagical | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 50p box|year=1950|num-b=41|num-a=43}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:21, 14 January 2023
Problem
The equation 
is satisfied when 
is equal to:
![$\textbf{(A)}\ \infty \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt[4]{2} \qquad \textbf{(D)}\ \sqrt{2} \qquad \textbf{(E)}\ \text{None of these}$](http://latex.artofproblemsolving.com/f/a/c/faceac5e5369ac6f7869a0600488ed0d14885599.png)
Solution
Solution 1:
Taking the log, we get 
, and 
. Solving for x, we get 
, and 
Solution 2:
is the original equation. If we let 
, then the equation can be written as 
. This also means that 
, considering that adding one to the start and then taking that to the power of 
does not have an effect on the equation, since is infinitely long in terms of raised to itself forever. It is already known that from what we first started with, so this shows that 
. If 
, then that means that .
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
