I probably don't have to tell you that the other answer is nonsense, but I'll do it all the same, just in case... Half-life is defined as the time it takes for some radioactive substance to decay to half its original amount.
If it takes n half-lives for some substance with a starting amount of A to decay to a final amount of B, then
[tex]B = \dfrac{A}{2^n}[/tex]
1. If the half-life is x, then after 1 half-life, 5000 units of this substance decays to 2500 units. After another half-life, this decays to 1250. After another, this in turn decays to 625.
So 3 half-lives are required.
In terms of the equation above, we solve for n such that
[tex]625 = \dfrac{5000}{2^n} \implies 2^n = 8 \implies n=3[/tex]
since 2³ = 8.
2. Count how many times you halve 15,000 to end up with 3750:
15,000/2 = 7500
7500/2 = 3750
===> 2 half-lives
In other words,
[tex]3750 = \dfrac{15,000}{2^n} \implies 2^n = 4 \implies n=2[/tex]
3. After 1 half-life, you would end up with
12,000/2 = 6000
and after another half-life,
6000/2 = 3000
i.e.
[tex]\dfrac{12,000}{2^2} = \dfrac{12,000}4 = 3000[/tex]
4. After 1 half-life, you have
26,000/2 = 13,000
After 2 half-lives,
13,000/2 = 6500
After 3 half-lives,
6500/2 = 3250
After 4 half-lives,
3250/2 = 1625
i.e.
[tex]\dfrac{26,000}{2^4} = \dfrac{26,000}{16} = 1625[/tex]
5. If the half-life is 3 years, then 15 years = 3 half-lives.
In the last 3 years, 20,000 would have decayed to 10,000.
In the second-to-last 3 years, 40,000 would have decayed to 20,000.
In the first 3 years, 80,000 would have decayed to 40,000.
i.e.
[tex]\dfrac{x}{2^3} = 10,000 \implies x = 8\times10,000 = 80,000[/tex]
