# How many half-lives have passed if 87.5% of a substance has decomposed? How many if 99.999% has decomposed?

For 87.5% decomposition 1/8 is left.
(1/8) = (1/2)^3
Three half lives is the answer

Fot 99.999% decomposition, 10^-5 is left.
Solve 10^-5 = (1/2)^n for the number of half-lives, n. Take the log of each side of the equatiuon for a quick solution.

## To find out how many half-lives have passed if 87.5% of a substance has decomposed, we need to determine the fraction of the substance that remains.

Given that 87.5% of the substance has decomposed, the remaining fraction is 1 - 0.875 = 0.125.

We can represent this as (1/8), which is equivalent to (1/2)^3.

So, to find the number of half-lives that have passed, we solve the equation (1/2)^n = (1/8), where n represents the number of half-lives.

Taking the logarithm of both sides will allow us to solve for n quickly.

log[(1/2)^n] = log(1/8)
n * log(1/2) = log(1/8)

Using logarithmic properties, we know that log(1/2) is equal to -log(2):

n * (-log(2)) = log(1/8)
n = log(1/8) / -log(2)

Evaluating this expression will give us the answer.

To calculate how many half-lives have passed if 99.999% of the substance has decomposed, we follow a similar approach.

After 99.999% decomposition, 0.00001 (10^-5) of the substance remains.

We can represent this as (1/2)^n = 10^-5, where n is the number of half-lives.

To solve for n, we take the logarithm of both sides:

log[(1/2)^n] = log(10^-5)
n * log(1/2) = log(10^-5)

Again, using log properties, we know that log(1/2) is equal to -log(2):

n * (-log(2)) = log(10^-5)
n = log(10^-5) / -log(2)

Evaluating this expression will give us the number of half-lives that have passed.