How to find the root number

S

smartcookie1/2thetime

New member
Joined
Jan 28, 2021
Messages
4
An isotope of Caesium has a half-life of 30 years, which means it loses 50% of its atoms every 30 years. Suppose we start with 10 kg sample of Caesium.
Q1. Write a rule that will calculate the amount of Caesium after ANY number of years.

So first I figured that if you begin with 10 kg, that would be the beginning part of the equation. Then after that I figured, if it loses 50% then you would do 100-50 which is just 50, or 0.5. And then you'd square that to get the half-life. This is what I have:

f(x)= 10*0.5^x

But I know that this isn't correct because when you plug in a given number, say 30 you'd expect the answer to be 5, because half of 10 is 5. Instead though, I got this 9.313 number, which is nowhere near close. While I'm doing this, I'm talking to a friend, and apparently my x is in half lives instead of years, which is a problem, because I am supposed to be figuring out the amount Caesium. And then he says that I instead I should be looking for the number that multiplies by 30 times to get to 0.5. Which basically means you have to find the root number, I think. Like how the 2 root of 16 is 4^2 or 4*4. At first I was like, okay, divided 0.5 by 30, but I realize it wouldn't work like that because that would be the opposite of multiplying. After that I was thinking maybe square rooting, but that doesn't work either.

And this is where I am stuck. Hopefully, this makes as much sense in my head as it does on paper. So I guess what my question is, how do I find the root number without having to spend and obnoxious amount of time plugging in a bunch of random numbers.
 
smartcookie1/2thetime said:
An isotope of Caesium has a half-life of 30 years, which means it loses 50% of its atoms every 30 years. Suppose we start with 10 kg sample of Caesium.
Q1. Write a rule that will calculate the amount of Caesium after ANY number of years.

So first I figured that if you begin with 10 kg, that would be the beginning part of the equation. Then after that I figured, if it loses 50% then you would do 100-50 which is just 50, or 0.5. And then you'd square that to get the half-life. This is what I have:

f(x)= 10*0.5^x

But I know that this isn't correct because when you plug in a given number, say 30 you'd expect the answer to be 5, because half of 10 is 5. Instead though, I got this 9.313 number, which is nowhere near close. While I'm doing this, I'm talking to a friend, and apparently my x is in half lives instead of years, which is a problem, because I am supposed to be figuring out the amount Caesium. And then he says that I instead I should be looking for the number that multiplies by 30 times to get to 0.5. Which basically means you have to find the root number, I think. Like how the 2 root of 16 is 4^2 or 4*4. At first I was like, okay, divided 0.5 by 30, but I realize it wouldn't work like that because that would be the opposite of multiplying. After that I was thinking maybe square rooting, but that doesn't work either.

And this is where I am stuck. Hopefully, this makes as much sense in my head as it does on paper. So I guess what my question is, how do I find the root number without having to spend and obnoxious amount of time plugging in a bunch of random numbers.
Click to expand...
You write:

f(x)= 10*0.5^x

What does 'x' represent?​
​
What does 'f(x)' represent?​
 
You're on the right track.

Let's say our function is of the form

[MATH]f(x) = A \alpha^x[/MATH]
[MATH]f(0) = A[/MATH]
We know that

[MATH]\dfrac{f(0)}{f(30)} = \dfrac{1}{\alpha^{30}} = \dfrac 1 2\\ \alpha^{30} = 2\\ \alpha = 2^{1/30}\\ f(x) = A(2^{1/30})^x = A 2^{x/30} [/MATH]
We can also write

[MATH]f(x) = A e^{(\ln(2)x)/30}[/MATH]
The two forms are equivalent.
 
Please be careful. A * 2^x/30 is not the same as A * 2^(x/30).
 
You must log in or register to reply here.
Top