find how many digits in the base 10 number 9^(9^9) has when written in base nine

Guest, Your answer does not contradicted mine :)

This was the question:

Find how many digits in the base 10 number 9^(9^9) has when written in base nine

Your comment was

So, the number of digits =369,693,099 + 1 =369,693,100 digits - in base 10 

YES I agree,

This whole thing is in base 10

What I said is that the answer is 

it will have 9^9+1 digits = 387 420 490 digits (base 10) BUT this is for the base 9 answer.

To put it slightly differently:

The base 9 conversion of 9^(9^9) will have 387 420 490 digits, with the number of digits expressed in base 10,

If the number of digits was expressed in base 9 it would look bigger.

I can do the conversion if you want:

387420490/9 = 43 046 721 R1

43046721/9=     4 782 969 R0

4782969/9  =        531 441 R0

531441/9  =            59049  R0

59049/9 =                6561   R0

6561/9   =                 729    R0

729/9     =                   81    R0

81/9       =                     9    R0

9/9         =                      1   R0

1/9        =                       0   R1

So the number of digits base 9   is  1 000 000 001

This certainly is no surprise I expected it to be a 1 at the beginning and at the end with zeros in the middle. 

10^(10^10) base 10 = 10^10 000 000 000 = 1 followed by 10 000 000 000 zerso = 10 000 000 001 digits

I just discovered the pattern ...

2^(2^2) = 2^4 = 10000(base2) :     5 digits long base 10 and 5 base 10 = 101 base 2

3^(3^3)= 3^27 = 1 followed by 27 zeros in base 3 :    28 digits long base 10 which is 1001  base 3

4^(4^4) = 4^ 256 = 1 followed by 256 zeros base 4:   257 digits long base 10 which is 10001 base 4

...

by extension:

9^(9^9) base 10 will be 1 000 000 001 base 9 units long in base 9.  Which is exactly what I got when I did it the super long way!!!

9^(9^9) =369,693,100 base 10

=812,298,765 base 9 [I think !!!].

Find how many digits in the base 10 number 9^(9^9) has when written in base nine

it will have 9^9+1 digits = 387 420 490 digits (base 10)

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simple example ( I needed this to think through my answer)

2^(2^2) base 10 written in base 2

2^4=16=10000₂

So that is 2^2+1=5 digits base 10 

OR    5₁₀ = 101₂

Melody: 9^(9^9) =9^(387,420,489) =387,420,489 x log(9) =369,693,099.631........etc.

So, the number of digits =369,693,099 + 1 =369,693,100 digits - in base 10

See Wolfram/Alpha Here:

http://www.wolframalpha.com/input/?i=9%5E9%5E9

The above answer has it in base 10 however we need it to be in base 9. 

9^(9^9) in base 10 =1,000,000,000 in base 9

Nice, Melody !!!