Question

fill in the blank 25 points
2 milligrams remain of an original 128 - milligram sample after 24 days.
$a = p(\frac{1}{2})^{\frac{t}{h}}$
a) what is the half - life of the sample? type your answer_ days
b) determine the quantity that remains after 24 days. type your answer_ mg

Explanation:

Step1: Substitute given values into formula

We know that $A = 2$ (remaining amount), $P=128$ (initial amount), and $t = 24$ (time). The formula is $A = P(\frac{1}{2})^{\frac{t}{h}}$. So, $2=128(\frac{1}{2})^{\frac{24}{h}}$.

Step2: Simplify the equation

First, divide both sides by 128: $\frac{2}{128}=(\frac{1}{2})^{\frac{24}{h}}$. Since $\frac{2}{128}=\frac{1}{64}$, and $\frac{1}{64}=(\frac{1}{2})^6$. So, $(\frac{1}{2})^6 = (\frac{1}{2})^{\frac{24}{h}}$.

Step3: Solve for $h$

Since the bases are the same, we can set the exponents equal: $6=\frac{24}{h}$. Cross - multiply gives $6h = 24$, and then $h = 4$.

Step4: Answer part b

We are already given in the problem that the quantity remaining after 24 days is 2 mg.

Answer:

a) 4
b) 2