SOLUTION: Radon is a naturally occurring radioactive gas which can collect in poorly ventilated structures. Radon gas is formed by the decomposition of radium-226 which has a half life of 16

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Question 1198803: Radon is a naturally occurring radioactive gas which can collect in poorly ventilated structures. Radon gas is formed by the decomposition of radium-226 which has a half life of 1622 years. The half-life of radon is 3.82 days. Suppose a house basement contained 38 grams of radon gas when a family moved in. If the source of radium producing the radon gas is removed so that the radon gas eventually decays, how long will it take until there is only 6.8 grams of radon gas present?
Found 4 solutions by Theo, math_tutor2020, ikleyn, MathTherapy:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
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.

https://blog.prepscholar.com/natural-log-rules

Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

x = number of days
y = amount of radon leftover, in grams
a = starting amount of radon gas = 38 grams
H = half-life of radon gas = 3.82 days

Half-life equation
y = a*(0.5)^(x/H)
y = 38*(0.5)^(x/3.82)

We want to find out the value of x when there are y = 6.8 grams of radon gas leftover.

Use logarithms to help isolate the exponent.
The relevant log rule used will be log(A^B) = B*log(A) to help pull down the exponent and isolate it.

y = 38*(0.5)^(x/3.82)
6.8 = 38*(0.5)^(x/3.82)
6.8/38 = (0.5)^(x/3.82)
0.17894737 = (0.5)^(x/3.82)
Log(0.17894737) = Log( (0.5)^(x/3.82) )
Log(0.17894737) = (x/3.82)*Log(0.5)
3.82*Log(0.17894737) = x*Log(0.5)
x = 3.82*Log(0.17894737)/Log(0.5)
x = 9.48274032162061
x = 9.48

It takes approximately 9.48 days for there to be 6.8 grams of radon gas remaining.
Answer by ikleyn(52800)   (Show Source): You can put this solution on YOUR website!
.

How @Theo solves this problem,  is  WRONG  TEACHING.


The standard method for this formulation is shown in the post by @math_tutor2020.


It is short, straightforward and  EXCLUDES  100  tons of unnecessary calculations
that @Theo makes in his post. 

    +-------------------------------------------------------------------------+
    |           When the problem gives the-half life duration,                |
    |         the correct analytic solution of  a  decay problem              |
    |    MUST be started from writing exponential decay function base 1/2,    |
    | if the opposite is not requested by the problem directly and explicitly.|
    |                                                                         |
    |    So called "ekt-form" is IRRELEVANT in such cases, so avoid use it,   |
    |             if you do not want to appear incompetent.                   |
    +-------------------------------------------------------------------------+

See also my lesson on radioactive decay problems under the link
    - Radioactive decay problems
in this site.

You will find many similar  (and different)  solved problems there.


This problem should be and must be solved in  5 - 6 - 7  lines,  no more than that.



Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Radon is a naturally occurring radioactive gas which can collect in poorly ventilated structures. Radon gas is formed by the decomposition of radium-226 which has a half life of 1622 years. The half-life of radon is 3.82 days. Suppose a house basement contained 38 grams of radon gas when a family moved in. If the source of radium producing the radon gas is removed so that the radon gas eventually decays, how long will it take until there is only 6.8 grams of radon gas present?
Neither a dissertation nor a novel needs to be written, as one person did, in order to solve this.
If  is given as "a" time periods, then k, the DECAY CONSTANT =  
So, we get: 
 
                    Continuous GROWTH/DECAY formula: 
                                                    ---- Substituting 6.8 for A, 38 for , and - .1815 for k
                                                   
                                               ------ Converting to LOGARITHMIC (natural) form
Time it'll take the radon to reduce to 6.8 grams, or

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