How to solve these type of questions

AFTER seeing Chris' answer, I found the error in mine

85184 = 44^3

(44^3)^(44^3)   = 44^(3x44^3) = 44^ (255552) = 44^(2 x 127776) = (44^2)^127776 = 1936^127776

as Chris found !

oh thats a hard one man....

Ya, I'm in grade eight. With this question even my dad was connfuzzled! Resorting to this website as a last hope....

i am 13 as well

Factor   85184  = 2^6 * 11^3  = [ 2^2 *11] [2^2 *11] [2^2 *11]   = [ 2^2 * 11]^3

Factor  1936  = 2^4 * 11^2  = [2^2 *11] [2^2 * 11]=   [2^2 *11]^2

So

85184   = ( [2^2 *11]^2])^(3/2) =   1936^(3/2)

So we have

[ 1936^3/2] ^ 85184 = 1936^127,776

Nice job, Chris !    Looked at it for a while...didin't quite come up with that !

EP and CPhill: How in the world did you guys get:85,184^85,184 =1936^127, 776???!!!, when direct calculation of 85,184^85,184 =4.153186160584012410564019499453 x 10^419,987???.

Well, Mr. Blarney, they have shown their work. Now all you need is a brain – an active one, of course. Having a petrified pile of oatmeal in its place won’t do.

Here is an easier way using logarithms:

Write 85184^85184 as a power of 1936

.

\(\dfrac{85184*log(85184)}{log(1936)} =127776\\ \;\\ 85184^{85184} = 1936^{1227776}\)

Normally I wouldn’t bother presenting this for you, because it’s like trying to teach a pìg to sing, and doing that is a waste time and annoys the pìg. However, there may be a bright eleven-year-old who will understand this. It’s not that difficult, really--not for a bright eleven-year-old.

Thanks, GA....I didn't think about going the "log route"...that's certainly  more efficient  !!!

Oh well.....I guess we can teach some pigs to fall off a "log"...others???.....Mmmm.....