College Algebra Colorado State University

We were not given a starting quantity, so we could either make up a value or use an unknown constant to represent the starting amount. To show that starting quantity does not affect the result, let us denote the initial quantity by the constant \(a\text{.}\) Then the decay of Bismuth-210 can be described by the equation \(Q(d)=a(0.87)^{d}\text{,}\) where \(d\) is a number of days.

To find the half-life, we want to determine when the remaining quantity is half the original: \(\dfrac{a}{2}\text{.}\) We start by setting our function equal to this desired output: \(\dfrac{a}{2}=a(0.87)^{d}\text{.}\) Then we divide both sides of the equation by \(a\) to get \(\dfrac{1}{2}=0.87^{d}\text{.}\) We can then take the common log of both sides to get \(\log(\dfrac{1}{2})=\log(0.87^{d})\text{,}\) and then use the exponent property of logs to find \(\log(\dfrac{1}{2})=d\log(0.87)\text{.}\) Dividing to solve for \(d\) gives us \(d=\dfrac{\log(\dfrac{1}{2})}{\log(0.87)}\approx 4.977 \text{ days}\text{.}\)

This tells us that the half-life of Bismuth-210 is approximately 5 days.

Since the half-life is 30 years, after 30 years, half the original amount, 100 mg, will remain.

After 60 years, another 30 years have passed, so during that second 30 years, another half of the substance will decay, leaving 50 mg.

After 90 years, another 30 years have passed, so another half of the substance will decay, leaving 25 mg.

Since we are looking for an annual decay rate, we will use an equation of the form \(Q(t)=a(1+r)^{t}\text{.}\) We know that after 30 years, half the original amount will remain. Using this information, we can write the equation \(\dfrac{a}{2}=a(1+r)^{30}\text{.}\) Dividing both sides by a gives us \(\dfrac{1}{2}=(1+r)^{30}\text{.}\) Taking the 30th root of both sides gives us \((\dfrac{1}{2})^{\frac{1}{30}}=1+r\text{,}\) and then subtracting one from both sides gives us \(r=(\dfrac{1}{2})^{\frac{1}{30}}-1\approx -0.02284\text{.}\)

This tells us cesium-137 is decaying at an annual rate of 2.284% per year.

To find how old the bone is, we first will need to find an equation for the decay of the carbon-14. We could either use a continuous or annual decay formula, but opt to use the continuous decay formula since it is more common in scientific texts. The half life tells us that after 5730 years, half the original substance remains. Solving for the rate, \(\dfrac{a}{2}=ae^{5730r}\text{.}\) Dividing by a gives us \(\dfrac{1}{2}=e^{5730r}\text{,}\) and then taking the natural log of both sides gives \(\ln(\dfrac{1}{2})=\ln(e^{5730r})\text{.}\) Use the inverse property of logs on the right side to get \(\ln(\dfrac{1}{2})=5730r\text{.}\) Then we can divide by 5730 to see \(r=\dfrac{\ln(\frac{1}{2})}{5730}\approx -0.000121\text{.}\)

Now we know the decay will follow the equation \(Q(t)=ae^{-0.000121t}\text{.}\) To find how old the bone fragment is that contains 20% of the original amount, we solve for \(t\) so that \(Q(t) = 0.20a\text{.}\)

Our initial equation to set up should look like \(0.20a=ae^{-0.000121t}\text{.}\) Then we divide by \(a\) to get \(0.20=e^{-0.000121t}\text{,}\) and then take the natural log of both sides to get \(\ln( 0.20)=\ln(e^{-0.000121t})\text{.}\) The inverse property reduces the right hand side, giving us \(\ln(0.20)=-0.000121t\text{,}\) and we can divide to solve for \(t\) to find \(t=\dfrac{\ln( 0.20)}{-0.000121}\approx 13301 \text{ years}\text{.}\)

So, if only 20% of the original carbon-14 is remaing, we can conclude that the bone fragment is about 13,300 years old.

Defining \(t\) to be time in months, with \(t = 0\) corresponding to last month, we are given two pieces of data: last month, \((0, 300)\text{,}\) and this month, \((1, 360)\text{.}\)

From this data, we can find an equation for the growth. Using the form \(C(t)=ab^{t}\text{,}\) we know immediately \(a = 300\text{,}\) giving \(C(t)=300b^{t}\text{.}\) Substituting in \((1, 360)\text{,}\) we see \(360=300b^{1}\text{,}\) and so \(b\dfrac{360}{300}=1.2\text{.}\)

This gives us the equation \(C(t)=300(1.2)^{t}\)

To find the doubling time, we look for the time when we will have twice the original amount, so when \(C(t) = 600\text{.}\) Our starting equation is \(600=300(1.2)^{t}\text{.}\) Dividing by 300 gives us \(2=(1.2)^{t}\text{,}\) and we can then take the common log of both sides to see \(\log(2)=\log(1.2^{t} )\text{.}\) The exponential property gives us the equation \(\log(2)=t\log(1.2)\text{,}\) and we can divide both sides by \(\log(1.2)\) to get \(t=\dfrac{\log(2)}{\log(1.2)}\approx 3.802\text{.}\) So it will take about 3.802 months for the number of cancer cells to double.

We can use the doubling time to find a function that models the number of site users, and then use that equation to answer the question. While we could use an arbitrary \(a\) as we have before for the initial amount, in this case, we know the initial amount was 120,000.

If we use a continuous growth equation, it would look like \(N(t)=120e^{r}t\text{,}\) measured in thousands of users after \(t\) months. Based on the doubling time, there would be 240 thousand users after 5 months. This allows us to solve for the continuous growth rate: \(240=120e^{5r}\text{,}\) so \(2=e^{5r}\text{.}\) Then, \(\ln(2)=5r\text{,}\) and so \(r=\dfrac{\ln(2)}{5}\approx 0.1386\text{.}\)

Now that we have an equation, \(N(t)=120e^{0.1386t}\text{,}\) we can predict the number of users after 12 months: \(N(12)=120e^{0.1386(12)}=633.140 \text{ thousand users}\text{.}\)

So after 1 year, we would expect the site to have around 633,140 users.

The point \(P\) appears to be half way between -2 and -1 in log value, so we should expect \(P\approx 10^{-1.5}=0.0316\text{.}\)

Since \(\log(6000)\approx 3.8\text{,}\) this point would belong on the log scale about here:

The value \(B\) corresponds to \(10^{2}=100\text{,}\) and the value \(C\) corresponds to \(10^{5}=100000\text{.}\)

The relative change is \(\dfrac{100,000}{100}=1000=10^{3}\text{.}\) The log of this value is 3. Thus, \(C\) is three orders of magnitude greater than \(B\text{,}\) which can be seen on the log scale by the visual difference between the points on the scale.

This tells us the first earthquake has about \(10^{9}\) times more earth movement than the baseline measure.

Doing the same with the second earthquake, \(S_{2}\text{,}\) with a magnitude of 8.0, the equation \(8.0=\dfrac{2}{3}\log(\dfrac{S_2}{S_0})\) gives us \(S_{2}=10^{12}S_{0}\text{.}\)

Comparing the earth movement of the second earthquake to the first, \(\dfrac{S_2}{S_1}=\dfrac{10^{12}S_0}{10^{9} S_0}=10^{3}=1000\text{.}\)

The second earthquake’s earth movement is 1000 times as large as the first earthquake.

Since the first quake has magnitude 3.0, \(3.0=\dfrac{2}{3}\log(\dfrac{S_1}{S_0})\text{.}\)

Since the second earthquake has twice as much earth movement, for the second quake, \(S_{2}=2\times10^{4.5}S_{0}\text{.}\)

Finding the magnitude, \(M=\dfrac{2}{3}\log(\dfrac{2\times 10^{4.5} S_0}{S_0})=\dfrac{2}{3}\log(2\times 10^{4.5})\approx 3.201\text{.}\)

The second earthquake with twice as much earth movement will have a magnitude of about 3.2.