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What is 2 tetrated -1 times?

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What is 2 tetrated -1 times?

Guest Dec 13, 2014

Best Answer

+5

From wikipedia:

(Limited) extension to negative heights[edit]

In order to preserve the original rule:

${^{(k+1)}a} = a^{({^{k}a})}$

for negative values of we must use the recursive relation:

${^{k}a} = \log_a \left( {^{(k+1)}a} \right)$

Thus:

${}^{(-1)}a = \log_{a} \left( {}^{0}a \right) = \log_{a} 1 = 0$

However smaller negative values cannot be well defined in this way because

${}^{(-2)}a = \log_{a} \left( {}^{-1}a \right) = \log_{a} 0$

which is not well defined.

Note further that for n = 1 any definition of $\,\! {^{(-1)}1}$ is consistent with the rule because

${^{0}1} = 1 = 1^n$ for any $\,\! n = {^{(-1)}1}$ .

Guest Dec 13, 2014

2⁺⁰ Answers

+23254

+5

I believe that, by definition ^-12 = log₂(⁰2) = log₂(1) = 0.

I would suggest going online to find a fuller explanation.

geno3141 Dec 13, 2014

+5

Best Answer

From wikipedia:

(Limited) extension to negative heights[edit]

In order to preserve the original rule:

${^{(k+1)}a} = a^{({^{k}a})}$

for negative values of we must use the recursive relation:

${^{k}a} = \log_a \left( {^{(k+1)}a} \right)$

Thus:

${}^{(-1)}a = \log_{a} \left( {}^{0}a \right) = \log_{a} 1 = 0$

However smaller negative values cannot be well defined in this way because

${}^{(-2)}a = \log_{a} \left( {}^{-1}a \right) = \log_{a} 0$

which is not well defined.

Note further that for n = 1 any definition of $\,\! {^{(-1)}1}$ is consistent with the rule because

${^{0}1} = 1 = 1^n$ for any $\,\! n = {^{(-1)}1}$ .

Guest Dec 13, 2014

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