Question

Will give a medal to the helper :) !!
equation is y=ae^ -0.00012t, a is amount object had to start, t is years
1.) has 32% remaining how old is it, graphically and algebraicly
2.) has 83% left how old "
3) what is the half life

Answer 1

Generally, if dealing with a half life problem, I'm used to using the formula:
Beginning Amount = Ending Amount * 2^n
where 'n' = (elapsed time / half life)
But to use that you'd have to know the half life to begin with.

Answer 2

k

Answer 3

So with that equation, y=ae^ -0.00012t,
do you know the beginning amount or the half-life?

Answer 4

no

Answer 5

I'm guessing that this is a calculus problem?

Answer 6

alg 2

Answer 7

I sure wish math problems would be stated in plain English!!!
I'll see what I can find. (I had to solve a problem like this a few months ago.)

Answer 8

k

Answer 9

Just wondering if that 'y' in that equation should be the Greek letter lambda λ ?

Answer 10

no

Answer 11

So, you are absolutely sure that the equation reads y=ae^ -0.00012t

Answer 12

yes

Answer 13

I'll check Wikipedia again. I could really help you with this IF they gave half-life , beginning amount, ending amount in ENGLISH.

Answer 14

okay

Answer 15

I wonder what whpalmer4 has to say

Answer 16

This isn't difficult once you have done a few of them.

\[ y=ae^{ -0.00012t}\]\(a = \)initial amount

1.) has 32% remaining how old is it, graphically and algebraicly

if it has 32% remaining, then \(y = 0.32a\) as \(a\) is the initial amount
\[0.32a = ae^{-0.00012t}\]\[0.32 = e^{-0.00012t}\]\[\ln 0.32 = -0.00012t\]you can do the rest.

2.) has 83% left how old "
same procedure, different numbers

3) what is the half life
the half life is the time to go from y = a to y = a/2
\[a/2 = ae^{-0.00012t}\]\[\frac{1}{2} = e^{-0.00012t}\]\[\ln \frac{1}{2} = -0.00012t\]

Answer 17

if you want to convert between the \[y = a_0(\frac{1}{2})^{-t/h}\] form where \(h\) is the half-life, and the form used here,
\[y = a_0e^{-\lambda t}\]just use \[h = \frac{\ln 2}{\lambda}\]

Answer 18

I guess that should be \[y = a_0(2)^{-t/h}\]Time to go to sleep!

Answer 19

Okay well thanks whpalmer4

Answer 20

Like almost all of this stuff, it helps if you do the problems often enough to remember how they go...

Answer 21

as opposed to stating them in plain English

Answer 22

Nah, it was a perfectly reasonable problem, IMO. That said, if you aren't used to the decay constant formulation, it's going to be confusing. The decay constant formulation is more convenient for various calculations because of the use of \(e\) as the base.

Answer 23

whpalmer4 I appreciate the explanation. Think I'll copy your answer and save it. (seriously)