A researchr assistant made 160mg of a radioactive, and found that there was only 20mg left after 45hours.
a) What is the half-life of this substance?
b) Write a function to represent the amount of substance left after x hours
c) If the researcher kew he needed 100mg of the same substance 12 hours after he was able to make a fresh batch, how much should he initially make?
Please help.i have a feeling that something like this will b on my math exam
Please show some steps so I can understand how u get it.
Thanks a bunch I really appreciate it!:)
This is an exponential function of the form
y=ae^x
Substituting x,y values (20,45)into the equation,
20=160*e^(45k)
take logs on both sides and solve for k
k=-(1/45)ln(8)
=-0.04621
So the mathematical model becomes:
y(t)=y0*e^(-0.04621t)
t=time in hours
y0=initial mass
y(t)=mass at time t hours.
Half life (value of t when y(t)/y0=1/2)
is
0.5 = e^(-0.04621t)
Take logs on both sides,
-0.04621t = ln(0.5)
t=15
can u do a more simpler way...we never learnt log yet..
a) To find the half-life of the substance, we need to determine the time it takes for the amount to reduce by half. In this case, the initial amount of the substance is 160mg, and it reduced to 20mg after 45 hours.
To find the half-life, we can set up the following equation:
Initial amount * (1/2)^(time/half-life) = Final amount
Plugging in the values we have:
160 * (1/2)^(45/half-life) = 20
To find the half-life, we can rearrange the equation as follows:
(1/2)^(45/half-life) = 20/160
(1/2)^(45/half-life) = 1/8
Now, let's take the logarithm of both sides of the equation to solve for the exponent:
log((1/2)^(45/half-life)) = log(1/8)
Using the logarithmic property, we can bring down the exponent:
(45/half-life) * log(1/2) = log(1/8)
Dividing both sides by log(1/2):
45/half-life = log(1/8) / log(1/2)
Using the properties of logarithms, we know that log(x) / log(y) is equivalent to log base y of x. Therefore:
45/half-life = log base (1/2) of (1/8)
We can simplify the equation further:
45/half-life = log base (1/2) of (1/2^3)
45/half-life = log base (1/2) of (1/2^(45/half-life))
Since the log base a of a^b is equal to b (log base a of a), we get:
45/half-life = 3
To solve for the half-life, we can multiply both sides by half-life and divide by 45:
half-life = (45 * 3) / half-life
half-life^2 = 135
Taking the square root of both sides:
half-life = √135
half-life ≈ 11.62
So, the half-life of this substance is approximately 11.62 hours.
b) To represent the amount of substance left after x hours, we can use the formula:
Amount left = Initial amount * (1/2)^(x/half-life)
c) If the researcher needs 100mg of the substance 12 hours after making a fresh batch, we can use the amount left formula to determine the initial amount.
Plugging in the values we have:
100 = Initial amount * (1/2)^(12/11.62)
To solve for the initial amount, we can rearrange the equation as follows:
Initial amount = 100 / (1/2)^(12/11.62)
Evaluating the expression:
Initial amount ≈ 238.2
So, the researcher should initially make approximately 238.2mg of the substance.