Solve for x using logarithms or substitution method

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SUMMARY

The forum discussion centers on solving the equation x^(x^(x^(x^(x^(x^(x^(x^(x…)))…) = 2 using two primary methods: logarithms and substitution. Participants, including MarkFL, magneto, and topsquark, successfully demonstrated their solutions, with topsquark illustrating the logarithmic approach and magneto employing substitution. An honorable mention was given to springfan25 for a minor error in the logarithmic method. The discussion highlights the effectiveness of both techniques in solving complex exponential equations.

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  • Understanding of exponential equations
  • Familiarity with logarithmic properties
  • Knowledge of substitution methods in algebra
  • Basic problem-solving skills in mathematics
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  • Study the properties of logarithms in depth
  • Practice solving exponential equations using substitution
  • Explore advanced techniques in algebraic manipulation
  • Review examples of complex equations solved with both methods
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Mathematics students, educators, and anyone interested in advanced algebraic techniques for solving exponential equations.

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Solve for x.

x^(x^(x^(x^(x^(x^(x^(x^(x…)))…) = 2
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I am standing in for Jameson this week. He has been working on various coding projects around the clock and so I am giving him a well-deserved break from his POTW duties.

Congratulations to the following members for their correct solutions:

1) MarkFL
2) magneto
3) anemone
4) topsquark
5) mente oscura
6) Pranav
7) eddybob123
8) jacobi

Honorable mention goes to springfan25 for having only made a minor but critical error in the last step of the logarithmic method.

There were basically two methods used by those who submitted solutions. One was to use logarithms, as illustrated by topsquark:

[math]x^{x^{x^x...}} = 2[/math]

[math]\ln \left ( x^{x^{x^x...}} \right ) = \ln(2)[/math]

[math]x^{x^{x^x...}} \cdot \ln(x) = \ln(2)[/math]

From the original problem statement [math]x^{x^{x^x...}} = 2[/math] so
[math]2 \ln(x) = \ln(2)[/math]

etc., so [math]x = \sqrt{2}[/math].

-Dan

The other was to use a substitution, as illustrated by magneto:

Let $p := x^{x^{x^{x^{\cdots}}}}$. We can rewrite the equation as $x^p = x^2 = 2$. Therefore, $x = \sqrt{2}$.
 

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