Half-Life

A

Angela123

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Oct 9, 2008
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Assume the substance has a half-life of 16 years and the initial amount is 143 grams.

How much remains at the end of 10 years?

How long will it be until only 20 % remains?

I would use the equation A=Pe^rt but that would just give me A=143e^16r. Where do I go from there?
 
Angela123 said:
Assume the substance has a half-life of 16 years and the initial amount is 143 grams.

How much remains at the end of 10 years?

How long will it be until only 20 % remains?

I would use the equation A=Pe^rt but that would just give me A=143e^16r. Where do I go from there?
Click to expand...

You need to use

A = P * 2[sup:lyypvi8t]-kt[/sup:lyypvi8t]

A/P = 2[sup:lyypvi8t]-kt[/sup:lyypvi8t]

1/2 = 2[sup:lyypvi8t](-k*10)[/sup:lyypvi8t]

Now find "k" and continue....
 
I really don't know how to get k. Could I divide both sides by 2? Then I would have .25=-10k then k=-.025. But what does k even stand for? This problem really confuses me.
 
Angela123 said:
I really don't know how to get k. Could I divide both sides by 2? Then I would have .25=-10k then k=-.025. But what does k even stand for? This problem really confuses me.
Click to expand...
You need a face-to-face tutor.
 
Angela123 said:
Well I don't have one. Is there any way you could explain it on here?
Click to expand...
Not really. The necessary background would have taken a chapter or two in your book. But if you'd like to try to do self-study, there are plenty of lessons available online:

. . . . .Google results for "logarithms"

. . . . .Google results for "log rules"

. . . . .Google results for "solving exponential equations"

Study at least two lessons from each link. Give yourself a few days to absorb the material. Then attempt the exercise, following the instructions provided earlier.

Eliz.
 
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