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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Solve for k. 2000 = 3500e^(k*2)

Frostbite
 4 years ago
Best ResponseYou've already chosen the best response.0Lets 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

Frostbite
 4 years ago
Best ResponseYou've already chosen the best response.0then 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So k = In(4/7)/2 is the final answer?

phi
 4 years ago
Best ResponseYou've already chosen the best response.0you can test your answer using a calculator

Frostbite
 4 years ago
Best ResponseYou've already chosen the best response.0I just toke it on the gefüle..

phi
 4 years ago
Best ResponseYou've already chosen the best response.0personally 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok cool! Thanks both of you!! :)
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