Question

A radioactive substance decays according to the following function, where yo is the initial amount present, and y is the amount present at time t (in days). y=yo e^-0.0936t Find the half-life of this substance. Do not round any intermediate computations, and round your answer to the nearest tenth days

Answer 1

Answer:

approximately 7.4 days

Step-by-step explanation:

Setting up the Problem

So we have the initial equation: $y = y_0*e^{-0.0936t}$ and we want to find how much time it takes to decay to the point where half of the substance is left.

So we can compare two points, and conveniently we can use the y-intercept, where: $y=y_0*e^0=y_0$

now to find the half life, we need to find where the y-value is half of $y_0$, and the only way this happens is if $y_0$ is multiplied by 0.5, meaning that $e^{-0.0936t}$ must be equal to 0.5, so let's solve for that

Solving for T

Now that we have some initial conditions set up, we just need to solve

$e^{-0.0936t}=0.5$

we can take the natural log of both sides:

$-0.0936t = \text{ln}(0.5)$

from here divide both sides by -0.0936

$t=\frac{\text{ln}(0.5)}{-0.0936}$

now from here we can approximate ln(0.5) using a calculator

$t\approx \frac{-0.69314718056}{-0.0936}$

now from here we can also approximate this using a calculator to get

$t\approx 7.40541859573$

rounding these to the nearest tenth we get:

$t\approx 7.4$