SOLUTION: My problem is with entering e raised to a negative power on my TI-84 Plus calculator. I know what answer I need to get and I can't figure out how to enter the problem correctly.

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: My problem is with entering e raised to a negative power on my TI-84 Plus calculator. I know what answer I need to get and I can't figure out how to enter the problem correctly.       Log On


   

Question 191254: My problem is with entering e raised to a negative power on my TI-84 Plus calculator. I know what answer I need to get and I can't figure out how to enter the problem correctly. I've looked on line and through my calculator manual to no avail!
The problem is: If 18 g of a radioactive substance are present initially and 3 yr later only 6 g remain, how much of the substance will be present after 11 yrs?
I get to the equation y=18e ^-.2310 x 7 and I know my answer is 5.02 grams, but how do I do this on my calc to get this answer? Thanks for any help!
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
 
A+=+Pe%5Ert

When t=0, A=18, plug those in:

A+=+Pe%5Ert
18+=+Pe%5Er%280%29
18+=+Pe%5E0
18+=+P%281%29
18+=+P

Substitute 18 for P in

A+=+18e%5Ert

When t=3, A=6, plug those in:

A+=+18e%5Ert
6+=+18e%5Er%283%29
6+=+18e%5E%283r%29

Divide both sides by 18

+6%2F18+=+%28+18%2Ae%5E%283r%29+%29%2F18+
1%2F3+=+e%5E%283r%29

Use principle A=e%5EN is equivalent to N=ln%28A%29

3r=ln%281%2F3%29

Divide both sides by 3

r+=+ln%281%2F3%29%2F3

In your TI-84 type  ln(1/3)/3 ENTER and get -.3662040962

r+=+-.3662040962+

Substitute -.3662040962 for r in

A+=+18e%5E%28rt%29

A+=+18e%5E%28-.3662040962t%29 

Now we substitute 11 for t

A+=+18e%5E%28-.3662040962t%29 
A+=+18e%5E%28-.3662040962%2811%29%29
A+=+18e%5E%28-4.028245058%29
A+=+18%28.0178055503%29
A+=+.3204999045

So after 11 years, there are only .3205 grams.

Edwin