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:

Instead, try using algebra:

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

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

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

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.