(9^9)^9
Haiwezi ikawa (9^9)^9 kwasababu
(9^9)^9 = (9)^81
kwahiyo jubu zuri litakuwa (9)^99
Ila mimi nitasema (9!)^99! au kama
hiyo utakataa basi hii 999!
hapana mkuu naona hujanielewa. ninavyosema 9^9 ninamaanisha ni 9 power 9. Hvyo ni sawa na kusema 9x9x9x9x9x9x9x9x9. Ningesema (9^2) hapo sawa ingekuwa 9x9=81. Hivyo hili ndo jibu sahihi kabisa kutokana na mtazamo wangu. (9^9)^9
Anyway,what about (9!^9!)^9!
Labda kama nimesahau hesabu lakini hebu tujikumbushe
a^a^a =[ (a^)^a]^a
= (a^)^axa
.
. . 9^9^9 = 9^9x9
whereby 9x9 is an exponent
= 81
Naweza kusahihishwa kama nimekosea
Anyway,what about (9!^9!)^9!
hapana, ninavyosema 9^9 ninamaanisha 9power9 na sio 9x9
Thanx mr cool,9! Implies [9times8times7times6times5times4times3times2times],shortly dat symbol means the factorial of a certain number, that means 9! is> than plain 9.
Proof that (9!)^99! > (9!^9!)^9!
since
99! = 99x98x97.......x9x8x7x6x5x4x3x2x1= 99x98x97.......x9!
= (99! - 9!)x9! then(9!)^99! = (9!)^(9!(99! - 9!) )
= (9!)^9!^(99! -9!)
= (9!^9!)^(99! -9!)let assume (9!^9!) = a
since
(9!)^99! = a^(99! -9!) and
(9!^9!)^9! = a^9!
then
(9!)^99! > (9!^9!)^9!
if
(9!^9!)^9! = (9!)^(9!x9!)
and
99! = (99!/9!)(9!)
then
(99!/9!) > (9!x9!)
Therefore
(9!)^99! > (9!^9!)^9!