Ok, homework time!

So, I have some chemistry homework. I know the formula for finding halflife decay is Y=A(1/2)^T/h

Where Y is the end amount, A is the starting amount, T is the time its been decaying, and H is the halflife.

What I need is a formula to find the actual halflife given the Starting and Ending amounts, and the time its been sitting.

Founder of UG's David Bowie Fan Club. Pm to join.

Founder of UG's "Rockers against being freakishly skinny" Club. PM to join.

Allow me to think upon that for a moment. misssssster freeman......
I knew I was going to get a halflife joke, but seriously, I need an answer here.
Founder of UG's David Bowie Fan Club. Pm to join.

Founder of UG's "Rockers against being freakishly skinny" Club. PM to join.
Y=A(1/2)^T/h into H=A(1/2)^T/Y?

EDIT: or not, just tested that
Founder of UG's David Bowie Fan Club. Pm to join.

Founder of UG's "Rockers against being freakishly skinny" Club. PM to join.
Last edited by zekk at Oct 17, 2008,
Quote by soulflyV
Re-arrange the equation, foo'.

'tis a hard equation to rearrange tho. i think it involves the natural logarithm of 2 tho....
Y=A(1/2)^(T/h)
Y/A = (1/2)^(T/h)

Y/A = 2^(-T/h)

log(base 2) (Y/A) = -T/h

there you go
Quote by Psychaberration
'tis a hard equation to rearrange tho. i think it involves the natural logarithm of 2 tho....

Uhh... I'm a sophomore in algebra 2, buddy. Getting a little ahead of where I am. I guess I could plug my givens into the equation and solve from there, but that leaves me with Time/Halflife as an exponent, and since I dont have the halflife, I'm stuck there.

{Like is i have 8grams of Cobalt-60, and over 10.5 years it reduces to 2grams, the equation I'm thinking of would be 2g=8g(1/2)^10.5/H Which leaves me stuck at 1g=8g^(10.5/H) which I'm not intierly sure is right because I'm guessing at math I've never done before)
Founder of UG's David Bowie Fan Club. Pm to join.

Founder of UG's "Rockers against being freakishly skinny" Club. PM to join.
i thought it was about the game

i cant wait til episode 3 comes out

you tricked me

Quote by jsbud11
Seriously if you leave UG without becoming a mod, I will kill someone.
Quote by Devopast
This is turning into fap-to-amazingfretman's-love-a-thon
Quote by romanqwerty
Y=A(1/2)^(T/h)
Y/A = (1/2)^(T/h)

Y/A = 2^(-T/h)

log(base 2) (Y/A) = -T/h

there you go

Explain the last formula to me, please. What is "log(base 2)"? This doesn't seem to answer my question, because my goal is to find the solution to H, and therefore find the actual halflife.
Founder of UG's David Bowie Fan Club. Pm to join.

Founder of UG's "Rockers against being freakishly skinny" Club. PM to join.
Quote by romanqwerty
Y=A(1/2)^(T/h)
Y/A = (1/2)^(T/h)

Y/A = 2^(-T/h)

log(base 2) (Y/A) = -T/h

there you go

To get the half life you would then go:

-T / (log(base 2) (Y/A) = H

Then just enter your values into the equation to find the half life.

I'm pretty sure you'll have to use change of base to do this equation as well.
Quote by zekk
Explain the last formula to me, please. What is "log(base 2)"? This doesn't seem to answer my question, because my goal is to find the solution to H, and therefore find the actual halflife.

log refers to the logarithm function.

Its basically how many powers of the base to i need to take to get to this number.

in this case

log(base 2) (Y/A) = -T/h

you take this logarthim of your ratio of initial amount to final amount. and equate it to the amount of times you have reached your half life.

For your question it is far simpler though.

You have 1/4 of the initial amount. Therefore you have reached exactly twice its halflife.
What exactly is "Base 2"? Keep in mind, I'm one marking period into algebra two here.
Founder of UG's David Bowie Fan Club. Pm to join.

Founder of UG's "Rockers against being freakishly skinny" Club. PM to join.
Im in australia, so i have no idea what algebra two is.

If you havn't learnt about logarithm's yet, look them up to understand what base 2 is.

Im guessing that you don't need to worry about a formula for that question. Your just supposed to realise that 2 grams is a quarter of 8 grams and that therefor you must have spent two halflifes of time. 1 to turn 8 into 4 and another to turn 4 into 2. Therefore 10.5 years is 2 halflifes and your halfife is 5.25years.
m(t) = m(0)e^(k*t) where m(t) is amount at time t, m(0) is initial amount.

Given a certain decay, solve for k, then set .5 = e^(k*t) and solve for t to find half life.
Schecter Loomis
LTD Horizon
Ibanez RGA121
Marshall DSL100
Peavey 5150

Quote by emagdnimasisiht
haha
This is the funniest thing i've ever read on UG.
lespaulrocks39, you sir are awesome.
Quote by lespaulrocks39
m(t) = m(0)e^(k*t) where m(t) is amount at time t, m(0) is initial amount.

Given a certain decay, solve for k, then set .5 = e^(k*t) and solve for t to find half life.

That doesn't answer how I solve for an exponent.
Founder of UG's David Bowie Fan Club. Pm to join.

Founder of UG's "Rockers against being freakishly skinny" Club. PM to join.
I'm basically in the same boat as the threadstarter so I guess I'll ask my question here too...we're doing half-life crap in physics and I've just barely started 2 Algebra so I don't know anything about logs.

Here's the equation, and I'm trying to solve for t:

10=32.313e^(-.0978t)

I know it has something to do with taking the ln of 10 and the exponent, but I don't know what happens to the 32.313e then.
Quote by romanqwerty
Im in australia, so i have no idea what algebra two is.

If you havn't learnt about logarithm's yet, look them up to understand what base 2 is.

Im guessing that you don't need to worry about a formula for that question. Your just supposed to realise that 2 grams is a quarter of 8 grams and that therefor you must have spent two halflifes of time. 1 to turn 8 into 4 and another to turn 4 into 2. Therefore 10.5 years is 2 halflifes and your halfife is 5.25years.

This man is speaking the truth...stop messing around with your silly and too-difficult-for-you equations.

And yes, I'm being serious

Have a nice day!
Quote by RoamingConflict
This one dream involved me, one random girl, midgets and a pie.

...and midgets ended up f*cking her. I got the pie.

Quote by zekk
That doesn't answer how I solve for an exponent.

Natural logarithms.
Schecter Loomis
LTD Horizon
Ibanez RGA121
Marshall DSL100
Peavey 5150

Quote by emagdnimasisiht
haha
This is the funniest thing i've ever read on UG.
lespaulrocks39, you sir are awesome.