Anyone want to help me with a math problem?

[Dynasty 10+ year member]

I **** at math //content.invisioncic.com/y282845/emoticons/*******.gif.a649d21efc0d1fd4890a6428166586c1.gif and could use some help on exponential growth and decay. Anyone want to give it a shot? An explanation could be helpful too.

A storage tank contains radium, a radioactive element with a half-life of 1600 years. Let f(t) represent the percentage of radium that remains in the tank at t years since the element was placed in the tank.

A) Find the equation for f(t).

B) What percentage of radium will remain in the tank after 100 years?

C) What percentage of radium iwill remain in the tank after 3200 years? Explain how you can find the result without using the equation in part A.

D) In approx how many years will there be 10% remaining?

Thanks.

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[Trey803 5,000+ posts]

hmm...that is actually a very simple math problem if you have the right equation which i dont know off the top of my head

 

[Twistid 5,000+ posts]

exponential growth + decay is easy dude

ti83/86 //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

f1 = formula

graph

awnser //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

 

[Calikid 5,000+ posts]

twistid is teh smartness

 

[bk12321 10+ year member]

f(x) = P * e^(rate*time)

P = principal amount, f(x) = final amount, e = a constant in the calculator.. its like 2.718 or something, r=rate and t=time

 

[.::DuD3::. 10+ year member]

one way that helped me remember that formula is pert, like the shampoo pert plus? never mind....

 

[Twistid 5,000+ posts]

hey ya thats pretty good dud3... never thaught of that

to bad last time we used exp growth + decay was in 11th grade

and geology 101 ... **** radioactive dating

 

[.::DuD3::. 10+ year member]

hey ya thats pretty good dud3... never thaught of that
to bad last time we used exp growth + decay was in 11th grade

and geology 101 ... **** radioactive dating

lol i know. acronyms are your friend.

 

[Dynasty 10+ year member]

f(x) = P * e^(rate*time)
P = principal amount, f(x) = final amount, e = a constant in the calculator.. its like 2.718 or something, r=rate and t=time

Is this right?

P=100

E= ?

R=1600

T= x

 

[fwb_1234 10+ year member]

you are gonna have to set up a system to get that R

set it up like this:

basic equation:

F(t) [amount remaining] = P[constant] * e[constant]^(r[constant]*t[variable])

when t=0:

1[as in 100% of all the radium] = Pe^(r*0)

e^0 = 1

P*1 = P

so in this case P = 1

so now you get : F(t) = 1e^(rt) = e^(rt)

second equation

when t=1600, F(t)= .5

so:

.5 = e^(r*1600)

take the natural log of both sides

ln(.5) = ln(e^r*1600))

--------

sidebar: ln(e^r*1600)) = (1600*r) * ln(e)

because ln(e) = 1

--------

ln.5 = 1600*r

thus:

r = ln(.5)/1600

therefore your final equation is:

F(t) = e^(ln(.5)/1600 * t)

Plug in those times and out comes the amount left in percentile form (example: .1 = 10%)

part C) not helping you

D) solve for t. you are going to use the log function (the "ln" button)