x^1/pi = 0 solve for x replied by 798 @ Fri, 25 Jan 2008 19:06:21 +0800

798 — Fri, 25 Jan 2008 19:06:21 +0800

Originally posted by HyuugaNeji:

ln 0 is undefined

tis part i quite confuse... cos i do not have matlab wif me.

x^1/pi = 0 solve for x replied by HyuugaNeji @ Fri, 25 Jan 2008 19:02:09 +0800

HyuugaNeji — Fri, 25 Jan 2008 19:02:09 +0800

Originally posted by 798:

x^-(1/pi) = 0
ln (x^-(1/pi)) = ln 0
(-1/pi)ln(x)=1
ln x = -pi
x = exp^-pi)

forget some of the natural log rules, hopefully i'm correct.

ln 0 is undefined

x^1/pi = 0 solve for x replied by 798 @ Fri, 25 Jan 2008 18:50:33 +0800

798 — Fri, 25 Jan 2008 18:50:33 +0800

x^-(1/pi) = 0
ln (x^-(1/pi)) = ln 0
(-1/pi)ln(x)=1
ln x = -pi
x = exp^-pi)

forget some of the natural log rules, hopefully i'm correct.

x^1/pi = 0 solve for x replied by manyu882 @ Fri, 25 Jan 2008 18:37:14 +0800

manyu882 — Fri, 25 Jan 2008 18:37:14 +0800

(x^-1)/pi=0

therefore, we can say that x^-1=0

which means 1/x=0

there for, x= infniti

when x approaches to 0

scenario 2

x^(-1/pi)=0

root (-1/pi) both sides.

therefore x=0.

x^1/pi = 0 solve for x replied by HyuugaNeji @ Wed, 23 Jan 2008 11:51:13 +0800

HyuugaNeji — Wed, 23 Jan 2008 11:51:13 +0800

Originally posted by popikachu:

TS means

x^(-1/pi)=0

answer,

(1/x)^(1/pi) = 0

1/( pi root(x) ) = 0

x = 0

which means x must be zero...

1/( pi root(x) ) = 0

1/0=pi root(x)

use your calculator and punch in 1 divided by 0.

x^1/pi = 0 solve for x replied by HyuugaNeji @ Wed, 23 Jan 2008 11:43:17 +0800

HyuugaNeji — Wed, 23 Jan 2008 11:43:17 +0800

x is undefined.

x is infinity only when 1/x approaches zero but not zero itself.

x^1/pi = 0 solve for x replied by ctstalin @ Sun, 20 Jan 2008 15:22:27 +0800

ctstalin — Sun, 20 Jan 2008 15:22:27 +0800

Originally posted by popikachu:

x^-1/pi = 0

x^-1 = 0

1/x= 0

1 = 0

i tink TS meant x^-(1/pi) = 0

dam...i totally forgotten wat to do...NS has made me stupid

x^1/pi = 0 solve for x replied by popikachu @ Sun, 20 Jan 2008 15:17:48 +0800

popikachu — Sun, 20 Jan 2008 15:17:48 +0800

TS means

x^(-1/pi)=0

answer,

(1/x)^(1/pi) = 0

1/( pi root(x) ) = 0

x = 0

which means x must be zero...

x^1/pi = 0 solve for x replied by eagle @ Sun, 20 Jan 2008 15:16:55 +0800

eagle — Sun, 20 Jan 2008 15:16:55 +0800

Originally posted by gobez14:

PAiseh paiseh the question supposed to be x^-1/pi = 0

x is infinity

x^1/pi = 0 solve for x replied by ^tamago^ @ Sun, 20 Jan 2008 15:16:02 +0800

^tamago^ — Sun, 20 Jan 2008 15:16:02 +0800

Originally posted by gobez14:

PAiseh paiseh the question supposed to be x^-1/pi = 0

not possible as x^(-1/pi) is always positive for all real values of x.

x^1/pi = 0 solve for x replied by gobez14 @ Sun, 20 Jan 2008 15:01:50 +0800

gobez14 — Sun, 20 Jan 2008 15:01:50 +0800

Originally posted by gobez14:

hmm but hor i sub 0 into x den 0^1/pi = math error on the calcualtor

PAiseh paiseh the question supposed to be x^-1/pi = 0

x^1/pi = 0 solve for x replied by gobez14 @ Sun, 20 Jan 2008 14:58:17 +0800

gobez14 — Sun, 20 Jan 2008 14:58:17 +0800

Originally posted by jay_rocks:

not 0 meh?

hmm but hor i sub 0 into x den 0^1/pi = math error on the calcualtor

x^1/pi = 0 solve for x replied by jay_rocks @ Sun, 20 Jan 2008 14:49:39 +0800

jay_rocks — Sun, 20 Jan 2008 14:49:39 +0800

not 0 meh?

x^1/pi = 0 solve for x replied by ^tamago^ @ Sun, 20 Jan 2008 14:45:08 +0800

^tamago^ — Sun, 20 Jan 2008 14:45:08 +0800

zero.

x^1/pi = 0 solve for x replied by gobez14 @ Sun, 20 Jan 2008 14:41:16 +0800

gobez14 — Sun, 20 Jan 2008 14:41:16 +0800

can any1 help with this qn?

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