HSC 2013 Maths Marathon (archive) (8 Viewers)

mahmoudali · Oct 16, 2013

Re: HSC 2013 2U Marathon

$\ \ $consider:$ 1 + \frac{1}{2} lnx + \frac{1}{4} (lnx)^2 + \frac{1}{8} (lnx)^3 +..... = \frac{2}{3} \ \ $find x$$

Sy123 · Oct 16, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
$\ \ $consider:$ 1 + \frac{1}{2} lnx + \frac{1}{4} (lnx)^2 + \frac{1}{8} (lnx)^3 +..... = \frac{2}{3} \ \ $find x$$

no soln

Shazer2 · Oct 16, 2013

Re: HSC 2013 2U Marathon

Well I thought you'd let the sum of geometric series = 2/3, but I don't know number of terms or x. I got this far:

$\frac {1(\frac {1}{2}lnx^{n}-1)}{\frac {1}{2}lnx-1}$

Yeah. If you give n, it's solvable I think.

Menomaths · Oct 16, 2013

Re: HSC 2013 2U Marathon

Shazer2 said:
Well I thought you'd let the sum of geometric series = 2/3, but I don't know number of terms or x. I got this far:

$\frac {1(\frac {1}{2}lnx^{n}-1)}{\frac {1}{2}lnx-1}$

Yeah. If you give n, it's solvable I think.

I think he wanted us to use limiting sum?

Shazer2 · Oct 16, 2013

Re: HSC 2013 2U Marathon

In that case, $x = \frac {1}{e}$

${S}_{\infty} = \frac {a}{1-r} \\ \\ \\ \indent \frac {1}{1-\frac {1}{2} \log _{e}{x}} = \frac {2}{3} \\ \\ \\ \indent 3 = 2-\log_{e}{x} \\ \\ \\ \indent e^{-1} = x \\ \\ \\ \indent \therefore x = \frac {1}{e}$

Menomaths · Oct 16, 2013

Re: HSC 2013 2U Marathon

Shazer2 said:
In that case, $x = \frac {1}{e}$

Yeah, that's what I got too

Menomaths · Oct 16, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
no soln

Is it wrong sai?

RealiseNothing · Oct 16, 2013

Re: HSC 2013 2U Marathon

Shazer2 said:
In that case, $x = \frac {1}{e}$

yer this is fine

mahmoudali · Oct 16, 2013

Re: HSC 2013 2U Marathon

what else would you use though
it writes +..... and i didnt write last term....
sy < me

Sy123 · Oct 16, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
Is it wrong sai?

yea lol there is no solution to the question you get ln(x) = -1, which is beyond even the 4U syllabus

RealiseNothing · Oct 16, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
yea lol there is no solution to the question you get ln(x) = -1, which is beyond even the 4U syllabus

wat?

Sy123 · Oct 16, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
what else would you use though
it writes +..... and i didnt write last term....
sy < me

I don't know what foreign mathematics you are going by

But in this country when we use ..... we mean infinite series

mahmoudali · Oct 16, 2013

Re: HSC 2013 2U Marathon

RealiseNothing said:
wat?

he's trying to troll its not working

Sy123 · Oct 16, 2013

Re: HSC 2013 2U Marathon

RealiseNothing said:
wat?

wait

oh my god

Sy123 · Oct 16, 2013

Re: HSC 2013 2U Marathon

dat fail

holy crap

mahmoudali · Oct 16, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
wait

oh my god

so he wasnt trolling he failed

DDDD

RealiseNothing · Oct 16, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
dat fail

holy crap

haha gg

Menomaths · Oct 16, 2013

Re: HSC 2013 2U Marathon

RealiseNothing said:
wat?

Sy123 said:
yea lol there is no solution to the question you get ln(x) = -1, which is beyond even the 4U syllabus

mahmoudali said:
what else would you use though
it writes +..... and i didnt write last term....
sy < me

So confused...Who to listen to =.=
Moey's out of the question so it leaves with realizenothing and sai =/

jkjk moey i can see you're good at maths

(relative to me at least)

mahmoudali · Oct 16, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
dat fail

holy crap

lnx = -1 (no soln)
how are you going to 1st 4U ?

Menomaths · Oct 16, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
dat fail

holy crap

...

(all i got to say)

HSC 2013 Maths Marathon (archive) (8 Viewers)

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