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C3 solving equation involving ln

http://pmt.physicsandmathstutor.com/download/Maths/A-level/C3/Papers-Edexcel/January%202015%20(IAL)%20QP%20-%20C34%20Edexcel.pdf

Hi for Q8 C in this paper, I don't now where I have gone wrong in my working, I am guessing its the bit highlighted. I have tried 2 methods using ln but all got me different answers. When I subbed x into e ^0.1t , I managed to get it right.

Where have I gone wrong in these two methods?
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answer for Method 2 should be 10/3 ln18/7

Thanks
(edited 6 years ago)
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Where you've gone wrong is in assuming that ln(a - b) = ln(a/b). The log rule you probably had in mind is ln(a) - ln(b) = ln(a/b), which doesn't apply here.

The examiners are probably expecting you to spot that the overall expression is a quadratic in e^(-0.1t). For clarity of working, you can put (say) p = e^(-0.1t), solve the quadratic for p, then reverse out the substitution to find the corresponding value of t.
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coconut64
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Original post by old_engineer
Where you've gone wrong is in assuming that ln(a - b) = ln(a/b). The log rule you probably had in mind is ln(a) - ln(b) = ln(a/b), which doesn't apply here. -> why ?

The examiners are probably expecting you to spot that the overall expression is a quadratic in e^(-0.1t). For clarity of working, you can put (say) p = e^(-0.1t), solve the quadratic for p, then reverse out the substitution to find the corresponding value of t.


I managed to get the answer at the end by solving the quadratic equation. So are you saying that you cannot use the ln rule to solve this question?

Thanks
Original post by coconut64
I managed to get the answer at the end by solving the quadratic equation. So are you saying that you cannot use the ln rule to solve this question?

Thanks


Yes, I am saying exactly that. I can't see any way of manipulating the expression into the form "ln(a) - ln(b)", so you can't bring that log rule into play.
(edited 6 years ago)
In method 1, you multiplied it all by e^(0.2t). But you forgot the 4e^(-0.1t)
Reply 5
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coconut64
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Original post by KappaRoss
In method 1, you multiplied it all by e^(0.2t). But you forgot the 4e^(-0.1t)


I did remember, it gives me 4e^(0.1t) -> no minus here

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