The function A=A0e^0.0099x models the amount in pounds in a particular radioactive material stored in a concrete vault, where x is the number of years since the material was put into the vault. If 400 pounds of the material are placed in the vault, how much time will need to pass for only 74 pounds to remain?

Question

e.mccormick · Answer

What part of this are you having trouble with?

anonymous · Answer

pretty much the whole thing. I'm not even sure where to start.

e.mccormick · Answer

OK.  They are giving you a formula there that should be some sort of continuous decay.
$A=A_0e^{0.0099x}$

Then they tell you that x is time, and that $A_0=400$.  What they want is the x when $A=74$.

rajee_sam · Answer

A is the final amount after x years and $$A _{0}$$ is the initial amount you start with. Just substitute the value in the function and solve for x

e.mccormick · Answer

And check that formula, make sure you copied it right.

anonymous · Answer

I missed a negative sign It's supposed to be $$A=A _{0}e ^{-0.0099x}$$

e.mccormick · Answer

That makes sense.  Because without the - it was an continuous growth formula.  With it, it is decay.

anonymous · Answer

so i set it up as $$74=400_{0}e ^{-0.0099x}$$?

e.mccormick · Answer

Well, the sub 0 is not needed on the 400, but other than that, yes.

anonymous · Answer

it was the sub 0 that was really messing me up I didn't know what to do with it

e.mccormick · Answer

OH!  It is this $A_n$ Ammount at time n.  So 0 is the start time.

e.mccormick · Answer

And the don't give you the n for the answer because that is what thay want you to solve for.

anonymous · Answer

oh! thanks!

e.mccormick · Answer

So what did you get?

anonymous · Answer

I'm not sure I did it right, I came up with 170.444

e.mccormick · Answer

$$74=400e ^{-0.0099x}\implies \
\frac{74}{400}=e ^{-0.0099x}\implies \
ln(\frac{74}{400})=ln(e ^{-0.0099x})\implies \
ln(74)-ln(400)=-0.0099x\implies \
\frac{ln(74)-ln(400)}{-0.0099}=x\implies \
170\approx x$$

anonymous · Answer

oh. I did do it right!

e.mccormick · Answer

Yep!  You did it right.

Subscripts have lots of meanings, like $(x_1,x_2)$ is another way of saying $(x,y)$.  In accounting type problems, P$_3$ would be the principal at year 3.  In Linear Algebra P$_3$ would be the space of 3rd degree polynomials.

anonymous · Answer

Thank you!

e.mccormick · Answer

np.  Have fun!