Question

what about x^x = 27

is it possible to find out that x=3 by some method?

Answer 1

Trial and error

Answer 2

xlogx = log27
xlogx = 3log3
x=3 ?
huh.. it's tough!

Answer 3

x^x=3^3
x=3

Answer 4

the only problem is that if you don't know 3^3 = 27, then what can be done?..... similarly, x^x = 287420498, what is x ?(surely that i know the answer but how can i find it if i don't know?)

Answer 5

ahhh,.... no no no log27 is not 3log9 and if you know xlogx=3log3 then you know the answer already...

Answer 6

Maybe what you can do it factorize the number into its prime factors, that's what I can see

Answer 7

xlogx=log27
xlogx=log(9*3)
xlogx=2log3+log3
xlogx=3log3
xlogx-3log3=0
3xlog(x/3)=0
log(x/3)=0
logx-log3=0
logx=log3
x=3

but it should be avoided and concluded from xlogx=3log3

Answer 8

:) i see ~~ then can i just write xlogx=3log3 directly instead of ''xlogx=log(9*3) => xlogx=2log3+log3'' ?

Answer 9

@Montreal: how did you get: xlogx-3log3=0
3xlog(x/3)=0

Answer 10

youre very right.

Answer 11

log x-log y= log(x/y) (if the logs have the same base)

Answer 12

right, so howz xlogx - 3log3 = 3xlog(x/3)

Answer 13

Its not right!

Answer 14

so how do you group logs with factors

Answer 15

xlogx-3log3
=log(x^x)-log(3^3)
=log(x^x/3^3)

This

Answer 16

right thanks