The​ half-life of a certain substance is 30 years. How long will it take for a sample of this substance to decay to ​80% of its original​ amount?

Question

Need some help lol

Answer 1

The half-life of the substance is 30 years.

Therefore after 1 half-life, only 1/2 of the substance will be left.

After 2 half-lives, only 1/4 = (1/2)² of the substance will be left.

After 3 half-lives, only 1/8 = (1/2)³ of the substance will be left.

The problem can now be stated: after how many half-lives will 80/100 of the substance be left?

Let's write this problem as a formula with a variable.

Let h be the number of half-lives after which 80/100 of the substance is left.

After h half-lives, we want only 80/100 = (1/2)^h to be left.

We can simplify this to 4/5 = (1/2)^h

To solve the above equation for h, we take the log_1/2 of both sides:

log_1/2 (4/5) = h

(We did that to bring the h down from the exponent.)

Let's write the variable first:

h = log_1/2 (4/5)

Changing log_1/2 to natural logarithms,

h = (ln (4/5)) / (ln (1/2))

If you don't like fractions, we can write it with whole numbers:

h = ((ln 4) - (ln 5)) / ((ln 1) - (ln 2))

h = ((ln 4) - (ln 5)) / (-ln 2) because the ln of 1 is 0

h is approximately 0.321928

Remember, 1 half-life is 30 years.

So approximately 0.321928 half-lives is approximately 0.321928 times 30 years = approximately 9.65784 years.

It will take 9.65784 years for the sample to decay to 80% of the original amount.

Answer 2

0.8 = 1 * (1/2)^(t/30)

log _(1/2) (0.8) = t/30

30 * log _(1/2) (0.8) = t

30 * log (0.8) / log (1/2) = t

9.6578428466 ≈ t years