My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
May 26th, 2015, 02:31 AM   #1
sum
Newbie
 
sum's Avatar
 
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.
sum is offline  
 
May 26th, 2015, 03:37 AM   #2
Senior Member
 
Joined: Apr 2014
From: Glasgow

Posts: 2,084
Thanks: 699

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Quote:
Originally Posted by sum View Post
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.
Benit13 is offline  
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.
Country Boy is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
good, question



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Good day! roketmike New Users 12 July 11th, 2014 12:58 PM
Do you have to be good at arithmetic to be a good at math? dthomas86 Elementary Math 11 September 2nd, 2011 12:22 PM
good one.... omerga Real Analysis 1 January 26th, 2009 10:04 AM





Copyright © 2018 My Math Forum. All rights reserved.