Question

New Problem:
Need Help Bad
A 100 lbs of radioactive substance decays to 35 lbs in 10 years. How much remains in 27 years?

I think if i could be taught how to find the value for Half-Life, that i could solve this problem.

Please help someone.

Answer 1

@phi if you are free to help i'd appreciate it.
@Titanic12 if you dont mind i really need help with this one part of this problem.

Answer 2

you could use the equaton
\[ A_t= A_0 \left(\frac{1}{2} \right)^{\frac{t}{\lambda}} \]
where A0 is the starting amount, t is the number of years, and lambda is the half-life

Answer 3

so its \[35=100(1/2)^{10/h}\]

Answer 4

they did not give the half-life, so we would have to figure it out, using the info they gave.

we could just use
\[ A_t = A_0 e^{kt} \]
where k is an unknown constant. We would have to solve for k using the info...

Answer 5

yes. can you find h ?

Answer 6

and the h being the half life, thats the part i cant get. im not sure how to find half life

Answer 7

cause i had gotten the same equation, this is the part i got stuck on

Answer 8

first divide both sides by 100
.35 = (1/2)^(10/h)

Answer 9

now take the log of both sides
write log "in front of" each side.

Answer 10

ok i got that
if im doing it right so far im to the point where this is the equation i have:
\[\log(0.35)=\frac{ -10 }{ h }\log(2)\]

Answer 11

ok

Answer 12

then i believe next step is to divide both sides by log(2) so
\[\frac{ \log(0.35) }{ \log(2)}=\frac{ -10 }{ h }\]

Answer 13

im not sure how to get the h by itself now

Answer 14

you could "flip" both sides

Answer 15

and then multiply both sides by -10

Answer 16

so it would be \[\frac{ \log2 }{ \log0.35 }=\frac{ h }{ -10 }\]
then \[-10[\frac{ \log(2) }{ \log(0.35) }]=h\]

Answer 17

and i just put that into my calculator to get a number?

Answer 18

yes
and it would be more convenient to find 1/h = k
so we can use
\[ A_t= A_0\left( \frac{1}{2}\right)^{kt} \]

Answer 19

ok so i put it in my calculator and i get h=6.602520221 or h=6.6

Answer 20

what do you mean it would be more convenient to find that? like i should use that equation instead of what i just did?

Answer 21

with exponents, it's always good to keep lots of digits.

when I suggest using k= 1/h that is because it's (slightly) simpler to do t*k rather than t/h.

Answer 22

ooh alright i see

Answer 23

so is that equation A = A(1/2)^kt the same as At=A0 e^kt?
just wrote it quickly without formatting it

Answer 24

related but different
if we used the second form (which does not show the half-life), the math would have been
0.35 = e^10k
ln(0.35) = 10 k
k= ln(0.35)/10

it's a bit faster to find the exponent.

Answer 25

oh alright let me write that out and see if i understand it

Answer 26

the "k" s will have different values for the two forms. In the first 1/k is the half-life
in the second, the half-life would be
ln(0.5)/k where k is ln(0.35)/10

Answer 27

ok yeah so i got h=6.6 again

Answer 28

can you finish the problem?

Answer 29

yes but just real quick, i did \[\frac{ \ln(0.5) }{ \frac{ \ln(0.35 }{ 10 } }\] but where is the ln(0.5) from?

Answer 30

sorry for all the quesitons, just trying to understand whats going on

Answer 31

all exponentials are related.
we use this property:
\[ e^{\ln(x)} = x\]
to write 1/2 as
\[ 0.5 = e^{\ln(0.5)} \]
and the original equation
\[ A_t= A_0 (0.5)^\frac{t}{h} \\
= A_0 \left( e^{\ln(0.5)}\right)^\frac{t}{h} \\
= A_0e^{\frac{\ln(0.5)}{h} \cdot t}
\]

Answer 32

oooh ok i see where that comes from now! thank you so much!!

Answer 33

i got my final answer of how many lbs after 27 years remains is 5.87471 lbs

Answer 34

pretty sure i did that right.

Answer 35

looks good

Answer 36

thank you so much phi, your a huge help!!