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How do you solve this? Solve for k. 2000 = 3500e^(k*2)

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Solve for k. 2000 = 3500e^(k*2)
Lets see: 2000 = 3500'e^(k*2) ln(2000) = ln(3500*e^(k*2)) ln(2000) = ln(3500) + ln(e^k*2) = ln(3500) + 2*k (ln(2000)-ln(3500))/2=k

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Other answers:

then to make it perhaps a bit more easy: (ln(2000)-ln(3500))/2=k (ln(2000)-ln(3500))/2=k <-> ln(2000/3500)/2=k k=ln(4/7)/2
So k = In(4/7)/2 is the final answer?
  • phi
you can test your answer using a calculator
I just toke it on the gef├╝le..
  • phi
personally I would start with \[ 2000 = 3500e^{2k} \] and divide both sides by 3500 and simplify \[ \frac{4}{7} = e^{2k} \] then take the natural log of both sides \[ \ln\left(\frac{4}{7}\right) = 2k\] divide both sides by 2 to get k
Ok cool! Thanks both of you!! :)

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