Question

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

Answer 1

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

Answer 2

@Frostbite @phi

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

you can test your answer using a calculator

Answer 7

I just toke it on the gefüle..

Answer 8

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

Answer 9

Ok cool! Thanks both of you!! :)