A 15-g sample of radioactive iodine decays in such a way that the mass remaining after t days is given by m(t) = 15e^(−0.051t), where m(t) is measured in grams. After how many days is there only 3 g remaining?
Solve 3 = 15 e^(−0.051t) for t
0.2 = e^(−0.051t)
ln (0.2) = -1.609 = -0.051 t
t = 31.6 (days)
Well, it seems like the radioactive iodine is going on a strict diet! To find out how many days it takes for there to be only 3 grams remaining, we can set up an equation.
We know that m(t) = 15e^(-0.051t), where m(t) represents the mass remaining after t days. We want to find the value of t when m(t) equals 3 grams. So, we can set up the equation:
3 = 15e^(-0.051t)
Now, let's do some "math"emagics to solve this equation:
3/15 = e^(-0.051t)
1/5 = e^(-0.051t)
To untangle this exponential equation, we can take the natural logarithm of both sides:
ln(1/5) = -0.051t
Now, let's do some more calculations:
t = ln(1/5) / -0.051
And when you plug this into a handy calculator or do some number crunching, you'll find that t ≈ 28.8084.
So, after approximately 28.8084 days, there will be only 3 grams of radioactive iodine remaining. Just long enough for it to show up on a scale and then make a swift exit.
To find the number of days after which there is only 3g remaining, we need to solve the equation m(t) = 3.
Given:
m(t) = 15e^(-0.051t)
m(t) = 3
We can set up the equation as follows:
15e^(-0.051t) = 3
To isolate the exponential term, divide both sides of the equation by 15:
e^(-0.051t) = 3/15
Simplifying the right side gives:
e^(-0.051t) = 1/5
To solve for t, take the natural logarithm (ln) of both sides of the equation:
ln(e^(-0.051t)) = ln(1/5)
Simplifying the left side using the property ln(e^x) = x:
-0.051t = ln(1/5)
Now, divide both sides of the equation by -0.051:
t = (ln(1/5)) / -0.051
Using a calculator, evaluate the right side:
t ≈ 54.8
Therefore, there will be only 3g remaining after approximately 54.8 days.
To find out after how many days there is only 3 g of the radioactive iodine remaining, we need to solve the equation m(t) = 3, where m(t) represents the mass remaining after t days.
The given equation for the mass remaining after t days is m(t) = 15e^(-0.051t).
Setting m(t) = 3, we have:
3 = 15e^(-0.051t)
To solve this equation for t, we will use logarithms.
First, divide both sides of the equation by 15:
3/15 = e^(-0.051t)
0.2 = e^(-0.051t)
Next, take the natural logarithm (ln) of both sides of the equation:
ln(0.2) = ln(e^(-0.051t))
Using the logarithmic property ln(e^x) = x, we simplify further:
ln(0.2) = -0.051t
Now, divide both sides of the equation by -0.051:
ln(0.2)/(-0.051) = t
Using a calculator, find the value of ln(0.2)/(-0.051):
t ≈ 76.8
Therefore, after approximately 76.8 days, there will be only 3 g of radioactive iodine remaining.