HELP HELP HELP ! attachment below .

Question

anonymous · Answer

attachment

anonymous · Answer

$$N=N_0e^{-kt}$$You have N_0 initially, which means that after the decay, you have$$N=0.70N_0$$
Substitute this and k into the equation to get
$$0.7N_0=N_0e^{-0.0001t}$$
Divide both sides by N_0$$0.7=e^{-0.0001t}$$
Take the natural log of both sides to solve for x.  

Let me know if you'd like me to go further with this.

anonymous · Answer

please go further :) i kind of understand .

anonymous · Answer

Take the natural log of both sides
$$\ln(0.7)=\ln(e^{-0.0001t})$$
The natural log and the act of raising e to a power are inverse operations.  This means that they will kind of "cancel" each other out, and we will be left with just the power.
$$\ln(0.7)=-0.0001t$$
Divide both sides by -0.0001 to get t by itself.
$$t=\frac{\ln(0.7)}{-0.0001}$$
This is a good answer, but it might be preferable to plug it into a calculator to get an approximate answer.

anonymous · Answer

3566.7?

anonymous · Answer

Looks good :).

anonymous · Answer

Oh wait, it says round to the nearest year.  3567

anonymous · Answer

it was incorrect :( oh man sorry bout that :( i should of gave you time to round .

anonymous · Answer

Sorry, I forgot that you posted the question as an attachment.