x^1/pi = 0 solve for x
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can any1 help with this qn?
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zero.
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not 0 meh?
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hmm but hor i sub 0 into x den 0^1/pi = math error on the calcualtorOriginally posted by jay_rocks:not 0 meh? -
PAiseh paiseh the question supposed to be x^-1/pi = 0Originally posted by gobez14:hmm but hor i sub 0 into x den 0^1/pi = math error on the calcualtor -
not possible as x^(-1/pi) is always positive for all real values of x.Originally posted by gobez14:PAiseh paiseh the question supposed to be x^-1/pi = 0 -
x is infinityOriginally posted by gobez14:PAiseh paiseh the question supposed to be x^-1/pi = 0 -
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... -
i tink TS meant x^-(1/pi) = 0Originally posted by popikachu:x^-1/pi = 0
x^-1 = 0
1/x= 0
1 = 0
dam...i totally forgotten wat to do...NS has made me stupid
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x is undefined.
x is infinity only when 1/x approaches zero but not zero itself. -
1/( pi root(x) ) = 0Originally 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...
1/0=pi root(x)
use your calculator and punch in 1 divided by 0. -
(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
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 undefinedOriginally 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. -
tis part i quite confuse... cos i do not have matlab wif me.Originally posted by HyuugaNeji:ln 0 is undefined