SOLUTION: I have only 3 questions like this The amount of a radioactive tracer remaining after t days is given by A = AO e-0.18t, where A0 is the starting amount at the beginning of the t

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: I have only 3 questions like this The amount of a radioactive tracer remaining after t days is given by A = AO e-0.18t, where A0 is the starting amount at the beginning of the t      Log On


   

Question 73053: I have only 3 questions like this
The amount of a radioactive tracer remaining after t days is given by A = AO e-0.18t, where A0 is the starting amount at the beginning of the time period. How much should be acquired now to have 40 grams remaining after 3 days?
47.9
48.8
61.6
68.6
please teach me through this. Thank you
Found 3 solutions by jim_thompson5910, bucky, ankor@dixie-net.com:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
This equation models the decay of a radioactive element. As time goes on the element decays and we want to know how much of the original amount A%5Bo%5D we should collect to have 40 grams after 3 days of decay. Since t represents the number of days, let t=3 and solve for A%5Bo%5D
40=A%5Bo%5De%5E%28-0.18%283%29%29Set the decay model equal to 40 (after 3 days it decays to 40 grams. If you plug in t=3, you get 40)Note: e is a constant
Divide both sides by e%5E%28-0.54%29
So
A%5Bo%5D=40%2Fe%5E%28-0.54%29
A%5Bo%5D=40%2F0.582748
A%5Bo%5D=40%2F0.582748
A%5Bo%5D=68.64030%0D%0AApproximately
So about 68.6 grams should remain after 3 days
Check:
A=%2868.64030%29e%5E%28-0.18%28t%29%29
If I let t=3 I should get A=40 (40 grams left over)
A=%2868.64030%29e%5E%28-0.18%283%29%29
A=%2868.64030%29%280.582748%29
A=39.9999975444%0D%0AWhich is really close to 40, so this answer works
Hope this helps.

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
A+=+A%5Bo%5D%2Ae%5E%28-0.18t%29
.
When t = 3 days, then you are told that A (the amount remaining) is 40 grams. Substitute
these values into the equation to get:
.
40+=+A%5Bo%5D%2Ae%5E%28-0.18%2A3%29
.
You need to solve this for A%5Bo%5D
.
First multiply out the exponent:
.
40+=+A%5Bo%5D%2Ae%5E%28-0.54%29
.
Finding e%5E%28-0.54%29 is just a calculator exercise. On most calculators the key will
be marked e%5Ex. Enter -0.54 and press the key to get 0.582748252. Substitute that
number into the equation to get:
.
40+=+A%5Bo%5D%2A%280.582748252%29
.
Now solve for A%5Bo%5D by dividing both sides of this equation by 0.582748252 to get:
.
A%5Bo%5D+=+40%2F0.582748252+=+68.64027449
.
So the answer is that you should start with 68.64027449 grams to have 40 grams left after
three days. Note that the closest answer in your list of answers is 68.6 grams.
.
Hope this work helps to get you familiar with solving exponential equations such as this.
This type of equation crops up in chemistry, physics, electronics, math, and societal studies
and you may need to understand how to handle them at some point.

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The amount of a radioactive tracer remaining after t days is given by A = AO e-0.18t, where A0 is the starting amount at the beginning of the time period. How much should be acquired now to have 40 grams remaining after 3 days?
:
Radio active decay formula should be given as:
A = Ao(e^-.18t)
:
A = 40; t = 3; Find Ao
:
40 = Ao(e^-.18*3)
40 = Ao(e^-.54
:
Using a good calculator; enter e^(-.54), should equal .5827482524
So you have:
.5827482524Ao = 40
:
Ao = 40/.5827482524
Ao = 68.64 grams to be acquired
:
If they want you to use natural logs, you add logs when you multiply:
ln(40) = ln(Ao) + ln(e^-.54)
ln(40) = ln(Ao) + (-.54)(ln(e), log equiv of exponents
:
Find the ln of 40. Remember the ln of e is 1, so we have:
3.688879454 = ln(Ao) - .54
:
3.688879454 +.54 = ln(Ao)
4.228879454 = ln(Ao)
:
Find the e^x; enter e^4.228879454:
Ao = 68.64, the same answer