Question 1198803
half life of radon gas is 3.82 days.
house basement contains 38 grams when the family moves in.
you want to know how long before the levels of radon gas are 6.8 grams.


formula you can use is f = p * e ^ (3.82 * r)
f is the future value
p is the present value
r is the interest rate per time period (days in this case).
n is the number of days


when f = 1 and p = 2, the formula becomes:
1/2 = e ^ (3.82 * r)
take the natural log of both sides of this equation to get:
ln(1/2) = ln(e^(3.82 * r))
since ln(e^(3.82 * r) = 3.82 * r * ln(e), you get:
ln(1/2) = 3.82 * r * ln(e)
since ln(e) = 1, this becomes:
ln(1/2) = 3.82 * r
divide both sides of the equation by 3.82 to get:
r = ln(1/2) / 3.82
solve for r to get:
r = -.1814521415


confirm by replacing r in the original equation and solving for f to get:
f = 2 * e ^ (-.181452415 * 3.82) = 1
this confirms the value of r is correct.


now that you know the value of r, you can solve the problem.
the formula for the problem becomes:
6.8 = 38 * e ^ (-.181452415 * t)
divide both sides of the equation by 38 to get:
6.8 / 38 = e ^ (-.181452415 * t)
take the natural log of both sides of the equation to get:
ln(6.8 / 38) = ln(e ^ (-.181452415 * t))
since ln(e ^ (-.181452415 * t)) = -.181452415 * t * ln(e), you get:
ln(6.8 / 38) = -.181452415 * t * ln(e)
since ln(e) = 1, this becomes:
ln(6.8 / 38) = -.181452415 * t
solve for t to get:
t = ln(6.8 / 38) / -.181452415 = 9.48274037


confirm by replacing t in the original equation and solving for f to get:
f = 38 * e ^ (-.181452415 * 9.48274037) = 6.8
this confirms the valjue of n is correct.


your solution is it will take 9.48274037 days for the amount of radon gas to reduce from 38 grams to 6.8 grams.


here are the natural logarithm rules that you should be aware of and that you can use to solve your logarithm problems.


<a href = "https://blog.prepscholar.com/natural-log-rules" target = "_blank">https://blog.prepscholar.com/natural-log-rules</a>