a certain radioactive isotope has a half life of approx 1,300 years .
How many years to the nearest year would be required for a given amount of this isotope to decay to 55% of that amount.
So I am not sure where to put the 1,300
1,300=2600*e^t
or A=x*e^1300
To determine the time required for a given amount of the isotope to decay to 55% of its original amount, we can use the formula for exponential decay:
A = A₀ * (1/2)^(t/T)
where:
A is the final amount (55% of the original amount),
A₀ is the initial amount (100% of the original amount),
t is the time passed, and
T is the half-life of the isotope.
In this case, we want to find the time (t) when the final amount (A) is 55% of the initial amount (A₀). We can rewrite the formula as follows:
0.55 = (1/2)^(t/T)
To solve for t, we can take the logarithm of both sides of the equation. Since we know the base of the exponential is 1/2, it is convenient to use the logarithm with base 2 (log₂) to cancel out the exponential.
log₂(0.55) = log₂((1/2)^(t/T))
Using the logarithmic property that log₂(a^b) = b * log₂(a), we have:
log₂(0.55) = (t/T) * log₂(1/2)
We want to solve for t, so we can rearrange the equation:
t/T = log₂(0.55) / log₂(1/2)
Now, we need to substitute T with its value, which is approximately 1,300 years.
t / 1,300 = log₂(0.55) / log₂(1/2)
To find the value of t to the nearest year, we can solve for t:
t = 1,300 * (log₂(0.55) / log₂(1/2))
Using a calculator, we can compute this expression:
t ≈ 1,300 * (-0.759 / -1) ≈ 989 years
Therefore, it would take approximately 989 years (to the nearest year) for the given amount of the isotope to decay to 55% of its original amount.