# Finding the Half Life

#### jwolfe890

##### New member
i'm trying to find the half life of a substance. i've seen some equations online, but when i carry them out the answer i get is different than the answer i get from using sites that calculate it for you.

for example, how would i find the half life of a substance where the initial quantity is 12, the remaining quantity is 3, and the time that elapsed to get to 3 is 2?

i've tried the one video i found on it and got a fraction instead of the right answer and would greatly appreciate any insight you have!

#### Dr.Peterson

##### Elite Member
i'm trying to find the half life of a substance. i've seen some equations online, but when i carry them out the answer i get is different than the answer i get from using sites that calculate it for you.

for example, how would i find the half life of a substance where the initial quantity is 12, the remaining quantity is 3, and the time that elapsed to get to 3 is 2?

i've tried the one video i found on it and got a fraction instead of the right answer and would greatly appreciate any insight you have!
Please show us the details: your work, your answer, the site you used to calculate it, the answer they gave, and the supposedly right answer you are judging by. Then we can help you figure out whether your work or that site was wrong! If the problem is from a textbook or similar source, quoting the entire problem may help, too. (That's what we tell you to do in the "Read before posting" instructions.)

#### Subhotosh Khan

##### Super Moderator
Staff member
i'm trying to find the half life of a substance. i've seen some equations online, but when i carry them out the answer i get is different than the answer i get from using sites that calculate it for you.

for example, how would i find the half life of a substance where the initial quantity is 12, the remaining quantity is 3, and the time that elapsed to get to 3 is 2?

i've tried the one video i found on it and got a fraction instead of the right answer and would greatly appreciate any insight you have!
What general equation would you expect this decaying reaction to follow?

#### jwolfe890

##### New member
What general equation would you expect this decaying reaction to follow?
honestly, i tried to reason it out but was having difficulty. i guess it would be in proportion to it at some rate.

Staff member

#### jwolfe890

##### New member
How did you get that?

original number 10, 5, 1

5/10 = 10b^1 / 10

1/2 ^ 1/1 = (b^1) ^ 1/1

.5 = b

#### Subhotosh Khan

##### Super Moderator
Staff member
original number 10, 5, 1

5/10 = 10b^1 / 10

1/2 ^ 1/1 = (b^1) ^ 1/1

.5 = b
Where did you get those numbers?

I see 12, 3 and 2

You have referenced good video - look at the full video and work through the problems solved there along with the presentation.

#### jwolfe890

##### New member
Where did you get those numbers?

I see 12, 3 and 2

You have referenced good video - look at the full video and work through the problems solved there along with the presentation.
10, 5, 1 is from a different number set

working through the video is how i got .5 instead of 1 in the first place though

#### Subhotosh Khan

##### Super Moderator
Staff member
10, 5, 1 is from a different number set

working through the video is how i got .5 instead of 1 in the first place though
half life of a substance where the initial quantity is 12, the remaining quantity is 3, and the time that elapsed to get to 3 is 2?
I am getting half-life = 1

#### jwolfe890

##### New member

I am getting half-life = 1

i think my mistake was in calculating (3/12)^1 wrong. I was using a Javascript function Math.pow(1/4, 1), which kept giving me .25, but the answer is actually just 1.

However, if the numbers are 8, 4, 2

(4 / 8) = (8b ^ 2) / 8

(1/2) ^ 1/2 = b ^ 2

0 = b

but b is supposed to equal 2??

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#### Subhotosh Khan

##### Super Moderator
Staff member
i think my mistake was in calculating (3/12)^1 wrong. I was using a Javascript function Math.pow(1/4, 1), which kept giving me .25, but the answer is actually just 1.

However, if the numbers are 8, 4, 2

(4 / 8) = (8b ^ 2) / 8 ............... From where did you get this equation? Show me step-by-step.

(1/2) ^ 1/2 = b ^ 2 .......................does not follow from above

0 = b

but b is supposed to equal 2??
.

#### jwolfe890

##### New member
beginning numbers 8, 4, 2

following along from the video equation.

4 = 8b^2

solve for b, thus:

4/8 = b^2

solve for b again:

(1/2) ^ 1/2 = (b^2) ^ 1/2

now from my understanding, this results in:

0 = b

#### stapel

##### Super Moderator
Staff member
beginning numbers 8, 4, 2

following along from the video equation.

4 = 8b^2

solve for b, thus:

4/8 = b^2

solve for b again:

(1/2) ^ 1/2 = (b^2) ^ 1/2

now from my understanding, this results in:

0 = b
The base cannot be zero. Sorry.

#### jwolfe890

##### New member
The base cannot be zero. Sorry.
((.5) ^ 1/2) = 0 though

what am i doing wrong?

#### stapel

##### Super Moderator
Staff member
How would I find the half-life of a substance where the initial quantity is 12, the remaining quantity is 3, and the time that elapsed to get to 3 is 2?
I think you mean something along the lines of the following:

You are given twelve grams of a radioactive substance. After two hours, you remeasure and find that the radioactive portion is now only three grams. Find the half-life of the substance.

There are various different ways to set this up. But here you've been given something extremely easy, because you can divide by two to find the amount left after one half-life, two half-lives, etc:

. . . . .t=0: 12

. . . . .t=?: 6

. . . . .t=2: 3

Clearly, as mentioned in a previous post, the half-life is one time unit (what I've called "hours" in my example).

If you want to use some half-life equation where you have to find the value of the base b, then:

. . . . .$$\displaystyle 3\, =\, 12\, (b)^2$$

. . . . .$$\displaystyle \dfrac{1}{4}\, =\, b^2$$

. . . . .$$\displaystyle \dfrac{1}{2}\, =\, b$$

original number 10, 5, 1

5/10 = 10b^1 / 10

1/2 ^ 1/1 = (b^1) ^ 1/1

.5 = b
I think this cryptic post is meant to signify something along the lines of the following:

You are given ten grams of a radioactive substance. After one hour, you find that five grams remain. Find the half-life of the substance.

After one hour, you have one-half left. So what must be the half-life?

If you want to use a half-life equation where you have to find the value of the base b, then:

. . . . .$$\displaystyle 5\, =\, 10\, b^1$$

. . . . .$$\displaystyle \dfrac{1}{2}\, =\, b$$

Check the answer by plugging it back in:

. . . . .$$\displaystyle 10\, \left(\dfrac{1}{2}\right)^1\, =\, 10\, (0.5)\, =\, 5$$

i think my mistake was in calculating (3/12)^1 wrong. I was using a Javascript function Math.pow(1/4, 1), which kept giving me .25, but the answer is actually just 1.
I have no idea where "3/12" (which equals 1/4) came from. Does it relate to any of these exercises?

beginning numbers 8, 4, 2
Are they to be interpreted as in the previous examples? If not, what do these numbers signify?

following along from the video equation.

4 = 8b^2

solve for b, thus:

4/8 = b^2

solve for b again:

(1/2) ^ 1/2 = (b^2) ^ 1/2

now from my understanding, this results in:

0 = b
How did you get that the square root of 1/2 was equal to zero? :shock:

. . . . .$$\displaystyle \dfrac{1}{2}\, =\, b^2$$

. . . . .$$\displaystyle \sqrt{\strut \dfrac{1}{2}\,}\, =\, \sqrt{\strut b^2\,}$$

. . . . .$$\displaystyle \dfrac{1}{\sqrt{\strut 2\,}}\, =\, b$$

Then rationalize the denominator.

By the way, were the base "b" to equal 2, then you'd have had:

. . . . .$$\displaystyle 8\, (2)^2\, =\, 8\, (4)\, =\, 32$$

...which doesn't work. On the other hand:

. . . . .$$\displaystyle 8\, \left(\dfrac{1}{\sqrt{\strut 2\,}}\right)^2\, =\, 8\, \left(\dfrac{1}{2}\right)\, =\, 4$$

...as desired.

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#### stapel

##### Super Moderator
Staff member
((.5) ^ 1/2) = 0 though

what am i doing wrong?
Since you haven't shown your steps, we have no idea what you might be doing wrong. Please reply with a clear listing of your steps. For instance, we know that the square root of zero is zero. How did you go from the square root of something NOT zero, and eventually end up with zero?

#### jwolfe890

##### New member
Since you haven't shown your steps, we have no idea what you might be doing wrong. Please reply with a clear listing of your steps. For instance, we know that the square root of zero is zero. How did you go from the square root of something NOT zero, and eventually end up with zero?

i listed the steps above:

beginning numbers 8, 4, 2

following along from the video equation.

4 = 8b^2

solve for b, thus:

4/8 = b^2

solve for b again:

(1/2) ^ 1/2 = (b^2) ^ 1/2

now from my understanding, this results in:

0 = b

#### Dr.Peterson

##### Elite Member
i listed the steps above:

beginning numbers 8, 4, 2

following along from the video equation.

4 = 8b^2

solve for b, thus:

4/8 = b^2

solve for b again:

(1/2) ^ 1/2 = (b^2) ^ 1/2

now from my understanding, this results in:

0 = b

Please show the details of your thinking on that last step: why do you think that (1/2)^(1/2) = 0?

Also, at one point you mentioned javascript. Does that mean that you are doing all of this in a program, rather than by hand?

#### jwolfe890

##### New member
Please show the details of your thinking on that last step: why do you think that (1/2)^(1/2) = 0?

Also, at one point you mentioned javascript. Does that mean that you are doing all of this in a program, rather than by hand?
eventually i'm going to try and do it in a program, but i can't even get it to work by hand right now.

i see i'm mistaken and it's actually 0.7071067811865476. i must misunderstand the equation because i'm supposed to be getting 2.

i thought there would be a simpler solution for this since it's a code challenge. like running the square root of numbers on top of the other equations in a JS function seems really cumbersome.