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What is 2 tetrated -1 times?

Guest Dec 13, 2014

Best Answer 

 #2
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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 k 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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2+0 Answers

 #1
avatar+17711 
+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
 #2
avatar
+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 k 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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