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dcfcviper · Oct 31, 2007, 06:31 AM

Hi All,

Looking for help re-arranging these formulae (making 'x' subject in both):

y=3+ [(4a+x)/(a-2x)]

and

ln(x^2 +1) -ln(x+1) =y

Thanks

asterisk_man · Oct 31, 2007, 06:44 AM

For the first one try starting by moving the 3 to the left and then multiplying both sides by (a-2x).

For the second one just remember that $e^{\ln x}=x$ so make each side the exponent of e and solve from there.

dcfcviper · Oct 31, 2007, 06:51 AM

Originally Posted by asterisk_man

For the first one try starting out by moving the 3 to the left and then multiplying both sides by (a-2x).

For the second one just remember that $e^{\ln x}=x$ so make each side the exponent of e and solve from there.

Thanks for the quick response. Done the second one now. Already started the first one a different way but have now got to the same line I got stuck on before.

asterisk_man · Oct 31, 2007, 07:18 AM

You can post your results or work and we can check it and give additional comments.

dcfcviper · Oct 31, 2007, 07:30 AM

y=3+ [(4a+x)/(a-2x)]

(y-3) = [(4a+x)/(a-2x)]

(y-3)(a-2x) = 4a+x

ay-2xy-3a+6x = 4a+x

ay-2xy-7a=-5x

asterisk_man · Oct 31, 2007, 10:22 AM

OK. What you have so far looks good. Just move all the terms that include x to the left and the terms that don't include x to the right. Then factor out the x from the one side and divide both sides by what's left.
Make sense?

dcfcviper · Oct 31, 2007, 10:27 AM

100% and so obvious no you've said it :D

Feel like a muppet now :(

Thanks for the help :thup:

asterisk_man · Oct 31, 2007, 10:39 AM

No problem. Don't forget to rate my answer if it was helpful ;)