Differential Equation Question Watch

v1oXx-
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#1
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Right I was doing a past paper, and I'm confused as how the mark scheme rearranged a certain equation. I got 5/6 marks available, but the final step lost me a mark.

How would I get from t=(1/k)ln(20/(20-kV)) to V = 20/k - (20/k)*e^-kt

I don't know how to use latex, sorry... I hope the above makes sense without it.
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alithegreat
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(Original post by v1oXx-)
Right I was doing a past paper, and I'm confused as how the mark scheme rearranged a certain equation. I got 5/6 marks available, but the final step lost me a mark.

How would I get from t=(1/k)ln(20/(20-kV)) to V = 20/k - (20/k)*e^-kt

I don't know how to use latex, sorry... I hope the above makes sense without it.

Write the whole log as one i.e. ln(20/20-kv)^(1/k) = t and then write e^t = (20/20-kv)^(1/k)

Then it should be easy, hope this helps
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v1oXx-
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I did that, but then what?
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rbnphlp
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(Original post by v1oXx-)
I did that, but then what?
from you original question write
kt=ln20-ln(20-kV) using log rules

-(kt-ln20)=ln(20-kV)

e^{-(kt-ln20)}=20-kV

Can you now rearrange for V?
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alithegreat
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(Original post by v1oXx-)
I did that, but then what?
Then multiply the indices be k on both sides i.e.

e^kt = (20/20-kv)^(1/k)*(k)

Then flip both sides to get e^-kt

e^-kt = (20-kv/20k)

Then it should be easy
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