May 26th, 2015, 01:31 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^21))/((x1)+1234+(log9999)) This is what I got. ((cos*pi)*ln*(e^x^21))/((x1)+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 02:45 AM. 
May 26th, 2015, 02:37 AM  #2  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,094 Thanks: 702 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
$\displaystyle f(x) = \frac{(\cos \pi) \ln \left( e^{x^2}  1 \right)}{ (x1) +1234 + \log (9999)}$ $\displaystyle = \frac{\ln\left(e^{x^2}  1\right)}{x+1233 + \log(9999)}$ 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 02:47 AM.  
May 28th, 2015, 05:40 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,976 Thanks: 807 
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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