- Thread starter
- #1

So, i set up 2 equations…

3=x(not)e^-k(1)

and 1=x(not)e^-k(3)

I know it is decaying ,but i don't know what i have to do with these equations that i made to find the value of k.

Thanks !

- Thread starter ISITIEIW
- Start date

- Thread starter
- #1

So, i set up 2 equations…

3=x(not)e^-k(1)

and 1=x(not)e^-k(3)

I know it is decaying ,but i don't know what i have to do with these equations that i made to find the value of k.

Thanks !

- Admin
- #2

You're off to a good start:

\(\displaystyle x_0e^{-k}=3\)

\(\displaystyle x_0e^{-3k}=1\)

I think what I would do next is solve both equations for $x_0$ and equate:

\(\displaystyle x_0=3e^{k}=e^{3k}\)

Next try dividing through by $e^k$ and then convert from exponential to logarithmic form.

- Thread starter
- #3

- Admin
- #4

You're welcome!Thanks!

I got k to be 0.549306144

and got a x(not) value of 5.196152423

I got it from here !

Thanks

I would get in the habit of obtaining/writing exact values rather than decimal approximations. I find:

\(\displaystyle k=\ln\left(\sqrt{3} \right)\)

\(\displaystyle x_0=3\sqrt{3}\)

I realize it is possible that you found these values and simply chose to write the approximations.