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# C3 June 2005 Question 7 watch

1. I've completed part 7a

I'm having trouble with part b, I have got the equation....

I think my steps are wrong.

2. it's more like

1850 = 336x/(1+0.12x), x= e^(0.2t)
1850 + 222x = 336x
x = 1850/114
0.2t = ln(1850/114)
etc.
3. 1850 = 2800(0.12)e^0.2t / 1+(0.12)e^0.2t
1850(1+0.12e^0.2t)=2800(0.12)e^0 .2t
1850+222e^0.2t=336e^0.2t
1850=(336-222)e^0.2t
1850=114e^0.2t
e^0.2t=(1850/114)
0.2t=ln(1850/114)
t=ln(16.228)/0.2
t=13.94
t=14 years
4. mark scheme here... http://x305.co.uk/PastPapers/Edexcel...%2005%20MS.pdf
5. (Original post by steve10)
it's more like

1850 = 336x/(1+0.12x), x= e^(0.2t)
1850 + 222x = 336x
x = 1850/114
0.2t = ln(1850/114)
etc.
(Original post by kitch91)
1850 = 2800(0.12)e^0.2t / 1+(0.12)e^0.2t
1850(1+0.12e^0.2t)=2800(0.12)e^0 .2t
1850+222e^0.2t=336e^0.2t
1850=(336-222)e^0.2t
1850=114e^0.2t
e^0.2t=(1850/114)
0.2t=ln(1850/114)
t=ln(16.228)/0.2
t=13.94
t=14 years

Thank you!

My working out is:

1850 = 336/0.12+e^-0.2t

1850(0.12+e^-0.2t) = 336

0.12 + e^-0.2t = 336/1850

e^-0.2t = 336/1850 - 0.12

0.2t = ln(336/1850 - 0.12)

t = - 13.93...

t = 14 years

Is it okay to do it this way? Also, can I ask why you both used 0.2t and not -0.2t?

Thanks again

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