A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
How do you solve this?
Solve for k.
2000 = 3500e^(k*2)
anonymous
 3 years ago
How do you solve this? Solve for k. 2000 = 3500e^(k*2)

This Question is Closed

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

Frostbite
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0So k = In(4/7)/2 is the final answer?

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

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

phi
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0Ok cool! Thanks both of you!! :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.