- Thread starter puremath
- Start date

Ok. I'll work on it.After working the first half-life problem, you should be able to do this one.

Is the formula needed k = [ln(1/2)]/(half-life)?After working the first half-life problem, you should be able to do this one.

I will target this problem later as I have all day to do math. No place to go. NYC is currently in lock down mode.No, check the previous threads.

After working the first half-life problem, you should be able to do this one.

For part B, must the first step be (0.80)(1.92 grams)?After working the first half-life problem, you should be able to do this one.

Please, explain your set up for B. I will solve for t as we will continue with exponential growth and decay throughout the remainder of the week in addition to starting trigonometry tomorrow.(a) \(\displaystyle A(100)=(2\text{ g})\left(\frac{1}{2}\right)^{\frac{100}{1620}}\approx1.91623115637718588\text{ g}\)

(b) \(\displaystyle \frac{1}{5}=\left(\frac{1}{2}\right)^{\frac{t}{1620}}\)

Solve for \(t\).

My answer for B makes no sense.(a) \(\displaystyle A(100)=(2\text{ g})\left(\frac{1}{2}\right)^{\frac{100}{1620}}\approx1.91623115637718588\text{ g}\)

(b) \(\displaystyle \frac{1}{5}=\left(\frac{1}{2}\right)^{\frac{t}{1620}}\)

Solve for \(t\).

Wow! I got it right. What does the value for t mean here?\(\displaystyle 5=2^{\frac{t}{1620}}\)

\(\displaystyle t=1620\log_2(5)\approx3761.523513717527\quad\checkmark\)

B. Find the time required for 80% of the 2 g sample to decay?

Wow! We will not be here for that.The unit of time here is years. With a half life of 1620 years, we know 75% will be gone in 2 half lives, or 3240 years, and 87.5% will be gone after 3 half-lives or 4860 years. So it seems reasonable that it would take some 3800 years for 80% to decay.