## swiftskier96 Group Title How do you solve this? Solve for k. 2000 = 3500e^(k*2) one year ago one year ago

1. swiftskier96

Solve for k. 2000 = 3500e^(k*2)

2. swiftskier96

@Frostbite @phi

3. Frostbite

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

4. Frostbite

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

5. swiftskier96

So k = In(4/7)/2 is the final answer?

6. phi

you can test your answer using a calculator

7. Frostbite

I just toke it on the gefüle..

8. 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

9. swiftskier96

Ok cool! Thanks both of you!! :)