+0  
 
0
318
2
avatar

What is 2 tetrated -1 times?

Guest Dec 13, 2014

Best Answer 

 #2
avatar
+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 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
 #1
avatar+17745 
+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

9 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.