Question 1201494
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x = number of hours that elapse
y = amount, in mg, of potassium-42 remaining
H = half-life of potassium-42 = 12.36 hours


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


We could use 1/2 in place of 0.5, but I find the 0.5 is easier to work with here.


Let's walk through a few scenarios for x.<ul><li>If one half-life elapses, then x = H. The exponent x/H becomes H/H = 1, meaning we have one copy of 0.5 multiplied to the starting amount 'a'. Half of the substance remains.</li><li>If x = 2H, two half-lives elapse, and x/H becomes 2H/H = 2. We'll have 2 copies of 0.5 multiplied with the 'a'. Therefore, (0.5)^2 = 0.25 = 25% of the substance remains. </li><li>If x = 3H, three half-lives elapse, and x/H becomes 3H/H = 3. We'll have 3 copies of 0.5 multiplied with the 'a'. Therefore, (0.5)^3 = 0.125 = 12.5% of the substance remains. </li></ul>And so on.


y = a*(0.5)^(x/H)
y = 500*(0.5)^(x/12.36)
This is one way to represent the exponential model, aka equation, aka function.


We have exponential decay because the base b = 0.5 fits the interval 0 < b < 1.


There are other formats to express this equation into.
Here are the steps to arrive at a second format.
y = 500*(0.5)^(x/12.36)
y = 500 * [ (0.5)^(1/12.36) ]^x
y = 500 * ( 0.94546361966231 )^x
The decimal value is approximate.
Round this value however your teacher instructs.


Here's yet another pathway to determine an equation.
Compare y = 500 * ( 0.94546361966231 )^x with y = 500*e^(kx) aka y = 500*(e^k)^x
We see that
e^k = 0.94546361966231
k = Ln(0.94546361966231)
k = -0.05607986897736
Therefore,
y = 500*e^(kx)
y = 500*e^(-0.05607986897736x)


So we have this candidate list of choices for the equation<ul><li>y = 500*(0.5)^(x/12.36)</li><li>y = 500 * ( 0.94546361966231 )^x</li><li>y = 500*e^(-0.05607986897736x)</li></ul>There are probably other formats to consider.


Personally I prefer the first format because I think it's easiest to work with. 
The half-life H = 12.36 is readily visible without having to do any calculations.


After we've determined the equation, we plug in x = 48 to figure out how much potassium-42 is left after 48 hours.
y = 500*(0.5)^(x/12.36)
y = 500*(0.5)^(48/12.36)
y = 33.8782897037247
y = 33.88
or you could say
y = 500 * ( 0.94546361966231 )^x
y = 500 * ( 0.94546361966231 )^48
y = 33.8782897037107
y = 33.88
or
y = 500*e^(-0.05607986897736x)
y = 500*e^(-0.05607986897736*48)
y = 33.8782897036631
y = 33.88
About 33.88 mg of potassium-42 remains after 48 hours.


An informal way to think about it:
1 half-life = 12.36 hrs
1 hr = 1/(12.36) half-lives 
48 hr = 48/(12.36) half-lives 
48 hr = 3.8835 half-lives approximately
The answer will involve a duration between 3 and 4 half-lives.


We start with 500 mg of potassium-42.<ul><li>After 12.36 hours (1 half-life), half of it goes away. We have 0.5*500 = 250 mg left.</li><li>After 24.72 hours (2 half-lives = 2*12.36 = 24.72 hrs), there will be 0.5*250 = 125 mg left.</li><li>After 37.08 hours (3 half-lives = 3*12.36 = 37.08 hrs), there will be 0.5*125 = 62.5 mg left.</li><li>After 49.44 hours (4 half-lives = 4*12.36 = 49.44 hrs), there will be 0.5*62.5 = 31.25 mg left.</li></ul>Therefore, the answer must be between 62.5 mg and 31.25 mg (corresponding to 3 and 4 half-lives).
This method won't nail down the exact value, but it does give an informal way to figure out where the answer should be located. 
It can serve as a way to check the answer.


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<font color=red>Summary:</font>


One equation is <font color=red>y = 500*(0.5)^(x/12.36)</font>
Other equations are possible. See above.


x = number of hours that elapse
y = amount, in mg, of potassium-42 remaining


<font color=red>Approximately 33.88 mg</font> of potassium-42 remains after 48 hours. 
Round this value however your teacher instructs.
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