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True or False? (1 Viewer)

Dreamerish*~

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loga(1) = 0, a > 1

True or false? :)
 
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Dreamerish*~

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SaHbEeWaH said:
false.
1 is a positive integer.
Oh my god, you're sharp.

Yes, I meant a>1. I'll change it now. Thanks for pointing that out! :eek:
 

Jago

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bah bastard.

loga1 = 0

a0 = 1

:)
 

Jago

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but doesn't 0/0 still equal 0?
 

tempco

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0/0 is an undefined quantity, not 0. afaik anyway.
 

who_loves_maths

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Originally Posted by Jago
but doesn't 0/0 still equal 0?
Originally Posted by nekkid
0/0 is an undefined quantity, not 0. afaik anyway.
we musn't forget that in the face of the Calculus branch of mathematics (even at the current HSC level), 0/0 is widely accepted to have the limiting value of 1.

it is incorrect and incomplete to generalise the statement that "0/0 is simply an undefined quantity" without the mention of possible limiting values. (of which 0 itself, other than 1, is also another)

Edit: in this situation, 0/0 is said to be indeterminate, rather than 'undefined'. the word 'indeterminate' in mathematics implies that there are more than one accepted or used value of, in this case, 0/0 depending on specific situations.
 
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Dreamerish*~

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who_loves_maths said:
as i mentioned in my last post, at the HSC level 0/0 is indeed defined. (as a limiting value)

the fact that Sin(x)/x = 1 as 'x' --> 0, depends on the definition that 0/0 =1
That's very impressive. Any money you'll ace 4U maths. :rolleyes:
 

who_loves_maths

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Originally Posted by Dreamerish*~
Oh my god, you're sharp.

Yes, I meant a>1. I'll change it now. Thanks for pointing that out!
Originally Posted by Slide_Rule
Actually, loga(1)=0 for all real a.
Originally Posted by Dreamerish*~
Except a = 1
in relation to the original question of this thread:

the logarithmic equation is, as Slide_Rule correctly points out, defined for all real 'a' {positive or negative} EXCEPT for a=0.

the value 0^0 is, like 0/0, indeterminate in mathematics and may carry different values depending on its use.
HOWEVER, there is very strong (but unofficial) argument in the mathematics community at large currently in support of the proposition 0^0 = 1 ; since this is the most 'useful' value that can be assigned to 0^0 via the stipulations of Calculus atm.
 

maths > english

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who_loves_maths said:
0/0 is widely accepted to have the limiting value of 1
when u get a situation <sup>f(x)</sup>/<sub>g(x)</sub> where f(x) and g(x) approach 0 as x approaches some value, the value of <sup>f(x)</sup>/<sub>g(x)</sub> isnt necessarily 1 because it depends on the manner in which f(x) and g(x) approach 0

e.g.

<sup>2<sup>x</sup>-1</sup>/<sub>x</sub> approaches log<sub>e</sub>2 as x approaches 0

<sup>x<sup>m</sup>-1</sup>/<sub>x<sup>n</sup>-1</sub> approaches <sup>m</sup>/<sub>n</sub> as x approaches 1

in the case above log<sub>a</sub>1 approaches 0 as a approaches 1


who_loves_maths said:
0/0 is said to be indeterminate, rather than 'undefined'. the word 'indeterminate' in mathematics implies that there are more than one accepted or used value of, in this case, 0/0 depending on specific situations
just to clarify im not saying who_loves_maths is wrong cause he did acknowledge that the value of <sup>0</sup>/<sub>0</sub> varies depending on the situation

<sup>0</sup>/<sub>0</sub> does very frequently turn out to be 1
 
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maths > english

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when analysing the <sup>0</sup>/<sub>0</sub> situation, using l'hopital's rule often helps

<sup>f(x)</sup>/<sub>g(x)</sub> as x approaches a equals <sup>f'(x)</sup>/<sub>g'(x)</sub> if f(x) and g(x) approach o or infinity as x approaches a
 
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maths > english

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who_loves_maths said:
there is very strong (but unofficial) argument in the mathematics community at large currently in support of the proposition 0^0 = 1
0<sup>0</sup> can be looked at as x<sup>x</sup> as x → 0

x<sup>x</sup> = e<sup>xlog<sub>e</sub>x</sup>

using l'hopital's rule, xlog<sub>e</sub>x → 0 as x → 0

:. x<sup>x</sup> as x → 0 goes to e<sup>0</sup> which equals 1

:. 0<sup>0</sup> can be interpreted as 1
 

maths > english

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who_loves_maths said:
^ hahaha... is this maths > english demonstrating his mathematical prowess? :D
point taken, i have sounded really stuck-up in this thread
 

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