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 May 26th, 2015, 02:31 AM #1 Newbie     Joined: Apr 2015 From: Russia Posts: 4 Thanks: 0 Math Focus: Algebra Good question This question is not very interesting, but I am not learning this in my maths (grade 9 maths) but I got it correct! I was very proud and wanted to share... Try to solve for yourself first... ((-cos*pi)*ln*(e^x^2-1))/((x-1)+1234+(log9999)) This is what I got. ((-cos*pi)*ln*(e^x^2-1))/((x-1)+1234+(log9999)) = (1*1)/(x+1233+4) = x+1237 I know this really isn't a difficult question, but I was really proud because I love maths and don't ever do maths like this PS forgive me for not knowing how to write this in latex form but I am inept. Last edited by skipjack; May 26th, 2015 at 03:45 AM.
May 26th, 2015, 03:37 AM   #2
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Quote:
 Originally Posted by sum This question is not very interesting, but I am not learning this in my maths (grade 9 maths) but I got it correct! I was very proud and wanted to share... Try to solve for yourself first... ((-cos*pi)*ln*(e^x^2-1))/((x-1)+1234+(log9999)) This is what I got. ((-cos*pi)*ln*(e^x^2-1))/((x-1)+1234+(log9999)) = (1*1)/(x+1233+4) = x+1237 I know this really isn't a difficult question, but I was really proud because I love maths and don't ever do maths like this PS forgive me for not knowing how to write this in latex form but I am inept.
Be careful....

$\displaystyle f(x) = \frac{(-\cos \pi) \ln \left( e^{x^2} - 1 \right)}{ (x-1) +1234 + \log (9999)}$
$\displaystyle = \frac{\ln\left(e^{x^2} - 1\right)}{x+1233 + \log(9999)}$

However,

$\displaystyle \ln \left( e^{x^2} - 1 \right) \approx x^2$

and

$\displaystyle \log(9999) \approx 4$

which yields

$\displaystyle f(x) \approx \frac{x^2}{x + 1237}$

But good on you for feeling good about solving problems! Just be careful to make sure you actually have the right answer before you celebrate.

Last edited by skipjack; May 26th, 2015 at 03:47 AM.

 May 28th, 2015, 06:40 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,919 Thanks: 785 By "e^x^2- 1" do you mean (e^(x^2))- 1 or e^(x^2- 1)? Those are very different! You appear to have interpreted it as e^(x^2- 1) and Benit13 as (e^(x^2))- 1. ln(e^(x^2- 1))= x^2- 1 but ln((e^(x^2))- 1) is not easily reducible.

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