After 150 thousand years, only 1/8 of the original amount of a particular radioactive waste will remain. The half-life of this radioactive waste is how many thousand years? 150,000 / 1/8 = 150,000 / .125 = 1,200,000 If this isn't right, then is the 1/8 supposed to stay as a fraction or be turned into a decimal or percent? Also, if this is not correct then what does the term "half-life" mean in the way this question is phrased?
Math Word Problem: After 150,000, only 1/8 of original amt remains...
- Thread starter Joystar77
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Joystar, this is an exponetial function. One way you can do this is by using an equation like:
\(\displaystyle N(t)=N_{0}e^{kt}\)
where:
\(\displaystyle N(t)\) is the amount remaining of the substance
\(\displaystyle N_{0}\) is the initial amount of the substance
\(\displaystyle k\) is the growth factor (if k > 0) or decay factor (if k < 0)
\(\displaystyle t\) is time
Usually for these problems \(\displaystyle k\) will not be given and you must first solve for \(\displaystyle k\) in order to solve for the main question. In order to solve for \(\displaystyle k\) you must be given the values of the three other variables. So we know that after 150,000 years (there's your \(\displaystyle t\)) there is 1/8 remaining of the substance. Even though you don't actually have the initial amount and the remaining amount, that is ok since you know that 1/8 remains, thus what would \(\displaystyle N(t)\) equal in terms of \(\displaystyle N_{0}\)?
See if you can solve for \(\displaystyle k\) then come back to us if you need any more help.
\(\displaystyle N(t)=N_{0}e^{kt}\)
where:
\(\displaystyle N(t)\) is the amount remaining of the substance
\(\displaystyle N_{0}\) is the initial amount of the substance
\(\displaystyle k\) is the growth factor (if k > 0) or decay factor (if k < 0)
\(\displaystyle t\) is time
Usually for these problems \(\displaystyle k\) will not be given and you must first solve for \(\displaystyle k\) in order to solve for the main question. In order to solve for \(\displaystyle k\) you must be given the values of the three other variables. So we know that after 150,000 years (there's your \(\displaystyle t\)) there is 1/8 remaining of the substance. Even though you don't actually have the initial amount and the remaining amount, that is ok since you know that 1/8 remains, thus what would \(\displaystyle N(t)\) equal in terms of \(\displaystyle N_{0}\)?
See if you can solve for \(\displaystyle k\) then come back to us if you need any more help.
HallsofIvy
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An "exponential" can be written with any base: if \(\displaystyle f(x)= a^x\) then \(\displaystyle f(x)= b^{log_b(a^x)}= b^{xlog_b(a)}\) so it's just a different coefficient. The "e" base is often used because it has nice "Calculus" properties which are not relevant here.
Because it says that "After 150 thousand years, only 1/8 of the original amount of a particular radioactive waste will remain" you can write \(\displaystyle P(t)= P(0)(1/8)^{t/150000}\). But \(\displaystyle 8= 2^3\), so this can be written as \(\displaystyle P(t)= P(0)(1/2^3)^{t/150000}= P(0)/2^{3t/150000}= P(0)/2^{t/50000}\). That is, because \(\displaystyle 8= 2^3\), the "half life" is 150000/3= 50000 years.
Because it says that "After 150 thousand years, only 1/8 of the original amount of a particular radioactive waste will remain" you can write \(\displaystyle P(t)= P(0)(1/8)^{t/150000}\). But \(\displaystyle 8= 2^3\), so this can be written as \(\displaystyle P(t)= P(0)(1/2^3)^{t/150000}= P(0)/2^{3t/150000}= P(0)/2^{t/50000}\). That is, because \(\displaystyle 8= 2^3\), the "half life" is 150000/3= 50000 years.
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Since you've posted this to "Arithmetic" (rather than to algebra or calculus), I'll guess that you're not supposed to use variables or formulas.
Yes, you're right about the meaning of "half-life": it's the amount of time it takes for only half to be left.
After 150,000 years, 1/8 is left. One half-life previously, how much (as a fraction) was left? (Hint: Multiply.) Another half-life previously, how much (as a fraction) was left?
Can you see a way to get the number of years that it took to get from whatever was the original amount down to only 1/2?
Math Word Problem Response
Serds: Thanks for letting me know this and what the definition mathematical term means.
Serds: Thanks for letting me know this and what the definition mathematical term means.
Math Word Problem Response
srmichael: I need to work this out step-by-step because then I get totally lost and confused. Is this first part right?
N = (No)e^(-kt)
where
No is the original amount
k is constant
t is time
substitute,
(1/8)*No = No * e^(-k*150000)
the No cancels:
1/8 = e^(-150000k)
ln (1/8) = -150000k
k = -150000/ ln(1/8)
after solving for k, substitute to this equation and get the value of t (half-life):
1/2 * No = No * e^(-kt)
Yes, we're looking for the half-life. Half-life is the time it takes for a substance to achieve 1/2 of its original amount when it undergoes radioactive decay. the formula used for radioactive decay is
N = (No)*e^(-k*t)
where
N = remaining amount after time, t
No = the original amount
k = some constant
t = time
since we don't know the k value (nothing is given), as well as the original amount (the No), express N in terms of No so we can cancel them on both sides. that's why the equation becomes
(1/8)*No = No * e^(-k*150000)
It is stated in the problem that after 150000 years (this is the variable t), the remaining amount will only be 1/8 of the original (the original is No, right?).
since the equation above contains No term, we can cancel them:
(1/8) = e^(-k*150000)
then get the natural log of both sides. That will cancel the e on the right side, leaving only the exponent:
ln (1/8) = ln (e^(-k*150000))
ln (1/8) = -150000k
k = -150000/ ln(1/8)
Get a calculator and solve the above equation to get k.
Is this so far correct?
srmichael: I need to work this out step-by-step because then I get totally lost and confused. Is this first part right?
N = (No)e^(-kt)
where
No is the original amount
k is constant
t is time
substitute,
(1/8)*No = No * e^(-k*150000)
the No cancels:
1/8 = e^(-150000k)
ln (1/8) = -150000k
k = -150000/ ln(1/8)
after solving for k, substitute to this equation and get the value of t (half-life):
1/2 * No = No * e^(-kt)
Yes, we're looking for the half-life. Half-life is the time it takes for a substance to achieve 1/2 of its original amount when it undergoes radioactive decay. the formula used for radioactive decay is
N = (No)*e^(-k*t)
where
N = remaining amount after time, t
No = the original amount
k = some constant
t = time
since we don't know the k value (nothing is given), as well as the original amount (the No), express N in terms of No so we can cancel them on both sides. that's why the equation becomes
(1/8)*No = No * e^(-k*150000)
It is stated in the problem that after 150000 years (this is the variable t), the remaining amount will only be 1/8 of the original (the original is No, right?).
since the equation above contains No term, we can cancel them:
(1/8) = e^(-k*150000)
then get the natural log of both sides. That will cancel the e on the right side, leaving only the exponent:
ln (1/8) = ln (e^(-k*150000))
ln (1/8) = -150000k
k = -150000/ ln(1/8)
Get a calculator and solve the above equation to get k.
Is this so far correct?
Math Word Problem Response
HallsofIvy: I get really confused with using log, the "e" base, and other Calculus properties. I have to do this step-by-step because if I don't then I get really lost and confused.
N = (No)e^(-kt)
where
No is the original amount
k is constant
t is time
substitute,
(1/8)*No = No * e^(-k*150000)
the No cancels:
1/8 = e^(-150000k)
ln (1/8) = -150000k
k = -150000/ ln(1/8)
after solving for k, substitute to this equation and get the value of t (half-life):
1/2 * No = No * e^(-kt)
Yes, we're looking for the half-life. Half-life is the time it takes for a substance to achieve 1/2 of its original amount when it undergoes radioactive decay. the formula used for radioactive decay is
N = (No)*e^(-k*t)
where
N = remaining amount after time, t
No = the original amount
k = some constant
t = time
since we don't know the k value (nothing is given), as well as the original amount (the No), express N in terms of No so we can cancel them on both sides. that's why the equation becomes
(1/8)*No = No * e^(-k*150000)
It is stated in the problem that after 150000 years (this is the variable t), the remaining amount will only be 1/8 of the original (the original is No, right?).
since the equation above contains No term, we can cancel them:
(1/8) = e^(-k*150000)
then get the natural log of both sides. That will cancel the e on the right side, leaving only the exponent:
ln (1/8) = ln (e^(-k*150000))
ln (1/8) = -150000k
k = -150000/ ln(1/8)
Get a calculator and solve the above equation to get k.
Is this so far correct?
HallsofIvy: I get really confused with using log, the "e" base, and other Calculus properties. I have to do this step-by-step because if I don't then I get really lost and confused.
N = (No)e^(-kt)
where
No is the original amount
k is constant
t is time
substitute,
(1/8)*No = No * e^(-k*150000)
the No cancels:
1/8 = e^(-150000k)
ln (1/8) = -150000k
k = -150000/ ln(1/8)
after solving for k, substitute to this equation and get the value of t (half-life):
1/2 * No = No * e^(-kt)
Yes, we're looking for the half-life. Half-life is the time it takes for a substance to achieve 1/2 of its original amount when it undergoes radioactive decay. the formula used for radioactive decay is
N = (No)*e^(-k*t)
where
N = remaining amount after time, t
No = the original amount
k = some constant
t = time
since we don't know the k value (nothing is given), as well as the original amount (the No), express N in terms of No so we can cancel them on both sides. that's why the equation becomes
(1/8)*No = No * e^(-k*150000)
It is stated in the problem that after 150000 years (this is the variable t), the remaining amount will only be 1/8 of the original (the original is No, right?).
since the equation above contains No term, we can cancel them:
(1/8) = e^(-k*150000)
then get the natural log of both sides. That will cancel the e on the right side, leaving only the exponent:
ln (1/8) = ln (e^(-k*150000))
ln (1/8) = -150000k
k = -150000/ ln(1/8)
Get a calculator and solve the above equation to get k.
Is this so far correct?
Math Word Problem Response
stapel: Actually, I wasn't sure what area to post it to because there isn't a section specifically stating "Math Word Problems", so I just thought it would go under Arithmetic. The course is actually Discrete Mathematics. I typed out what the full word problem says and it doesn't say anything at all about using variables or formulas. I have to work this out step-by-step with the variables, formulas, and words so that way I can understand the material. I have a learning disability which is Epilepsy so this causes me to lack comprehension and understanding. Sometimes, I hate these words in word problems because it confuses me.
No, I don't see how you come up with the fraction after the one-half life previously. The only fraction I see in the problem is 1/8. To my understanding, the word problem says, "After 150 thousand years, only 1/8 of the original amount of a particular radioactive waste will remain."
150,000 years
1/8 of original amount of particular radioactive waste will remain
The amount of time I know is 150,000 years. When it says 1/8 of the original amount of a particular radioactive waste will remain, then doesn't this mean that 1/8 is the original amount of a particular radioactive waste then shouldn't that mean just one will remain because of it saying "A particular"? Am I reading too much into the word problem and making it more difficult than what it really should be? No, I don't seem to understand or see a way to get the number of years that it took to get from whatever was the original amount down to 1/2.
stapel: Actually, I wasn't sure what area to post it to because there isn't a section specifically stating "Math Word Problems", so I just thought it would go under Arithmetic. The course is actually Discrete Mathematics. I typed out what the full word problem says and it doesn't say anything at all about using variables or formulas. I have to work this out step-by-step with the variables, formulas, and words so that way I can understand the material. I have a learning disability which is Epilepsy so this causes me to lack comprehension and understanding. Sometimes, I hate these words in word problems because it confuses me.
No, I don't see how you come up with the fraction after the one-half life previously. The only fraction I see in the problem is 1/8. To my understanding, the word problem says, "After 150 thousand years, only 1/8 of the original amount of a particular radioactive waste will remain."
150,000 years
1/8 of original amount of particular radioactive waste will remain
The amount of time I know is 150,000 years. When it says 1/8 of the original amount of a particular radioactive waste will remain, then doesn't this mean that 1/8 is the original amount of a particular radioactive waste then shouldn't that mean just one will remain because of it saying "A particular"? Am I reading too much into the word problem and making it more difficult than what it really should be? No, I don't seem to understand or see a way to get the number of years that it took to get from whatever was the original amount down to 1/2.
Since you've posted this to "Arithmetic" (rather than to algebra or calculus), I'll guess that you're not supposed to use variables or formulas.
Yes, you're right about the meaning of "half-life": it's the amount of time it takes for only half to be left.
After 150,000 years, 1/8 is left. One half-life previously, how much (as a fraction) was left? (Hint: Multiply.) Another half-life previously, how much (as a fraction) was left?
Can you see a way to get the number of years that it took to get from whatever was the original amount down to only 1/2?
You're welcome.