A radioactive substance decays at a rate of y = ae^-0.1483t, where t in in hours. Find the half-life of the substance.
How do I solve this problem if I am only given the constant.
so you want:
(1/2)a = ae^(-.1483t)
.5 = e^(-.1486t)
take ln of both sides and use rules of logs
-.1486t = ln .5
t = ln .5/-.1486 = appr 4.66 hours
To find the half-life of a substance, we need to determine the time it takes for half of the substance to decay.
In the given equation y = ae^-0.1483t, y represents the amount of the substance remaining at time t, and a is the initial amount at time t=0.
To find the half-life, we need to set y equal to half of the initial amount (a/2), and then solve for t.
So, we have:
a/2 = ae^-0.1483t
To solve for t, we can cancel out the common factor 'a' on both sides of the equation:
1/2 = e^-0.1483t
Next, we can take the natural logarithm (ln) of both sides of the equation to isolate the exponential term:
ln(1/2) = ln[e^-0.1483t]
Using the property ln(a^b) = b * ln(a), we can simplify the equation further:
ln(1/2) = -0.1483t * ln(e)
Since ln(e) equals 1, the equation becomes:
ln(1/2) = -0.1483t
Now, we can solve for t by isolating it on one side of the equation:
t = ln(1/2) / -0.1483
Using a calculator, we can evaluate ln(1/2) ≈ -0.693:
t ≈ -0.693 / -0.1483
t ≈ 4.678 hours
Therefore, the half-life of the substance is approximately 4.678 hours.