How to

Find Initial Amount and Amount After Time From Half-Life (y=ae^(kt))

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The half-life of palladium-100 is four days. After 12 days, a sample of palladium-100 has been reduced to a mass of six milligrams. What was the initial mass of the sample and what is the mass seven weeks after the start? To answer the questions, we’ll be using the exponential function shown below, where p of t is the amount after time t. P sub zero is the initial or starting amount. K is the exponential growth or decay rate expressed as a decimal and t is time.

We’ll begin by determining k using the fact that the half-life of palladium-100 is four days. Remember, the half-life is the time it takes for half of the mass to decay. So we know we know the time t in days is equal to four. And now to marl the half-life, let’s let p sub zero, the starting amount, be equal to two milligrams. So if p sub zero is equal to two milligrams and the half-life is four days, after four days, half of the two milligrams will be left, which means p of four is equal to one. Again, after four days, one milligram of mass is left. And now perform substitution into the equation and solve for k. On the left side, we have p of four, which is one equals p sub zero is two. E raised to the power of k times t, where k is unknown and t is four days.

The next step is isolate the exponential part by dividing both sides by two. Simplifying, we have 1/2 equals two divided by two simplifies to one. On the right side, we have e raised to the power of k times four or e raised to the power of four k. And now to solve for k, because k is in the position of the exponent and we have base e, we will take the natural log of both sides of the equation. This gives us natural log, 1/2 is equal to natural log of e raised to the power of four k. On the right side, we apply the power property of logarithms. Natural log of e raised to the power of four k is equal to four k times natural log e, and natural log e is equal to one, so this simplifies out. To solve for k, we divide both sides by four.

Simplifying on the right first, four divided by four simplifies to one, giving us k, equals on the right side, we have natural log 1/2 divided by four, but let’s write that as 1/4 natural log 1/2. And now sub divide of k into the exponential function. We have p of t equals p sub zero times e raised to the power of 1/4 natural log 1/2 times t. And now we can use this function with the information that after 12 days, a sample of palladium-100 has been reduced to a mass of six milligrams to determine the initial mass of the sample. Using the given information, time t is 12 days. The sample has been reduced to a mass of six milligrams after the 12 days, which means the amount after 12 days or p of 12 is equal to six milligrams, and we’re trying to find the initial mass, which is p sub zero. So again, we’ll substitute the known values into the exponential function. Well on the left side, we have p of 12, which is six, equals, on the right side, we don’t know p sub zero.

Then we have e raised to power of 1/4 natural log 1/2 times t, where t is time in days, which gives us an exponent of 1/4 natural log 1/2, and times t, which is 12. Let’s go and simplify the exponent here. Now this 1/4 times 12 is equal to three, which gives us six equals p sub zero times e, raised to the power of three natural log 1/2. Now to solve for p sub zero, we divide both sides by e raised to the power of three natural log 1/2. Simplifying on the right first, this quotient simplifies to one giving us p sub zero is equal to the quotient on the left, and now we go to the calculator. We have six divided by second natural log brings up e raised to the power of and the exponent is three natural log 1/2. If we want 0.5, close parenthesis and enter. Notice p sub zero is 48 milligrams, which means the initial mass of this sample is 48 milligrams, and let’s also place 48 for p sub zero in our exponential function.

We now know p of t is equal to 48 times e raised to the power of 1/4 natural log 1/2 times t. And now that we have our function in terms of t, we can answer the last question which is what is the mass in milligrams seven weeks after the start? We need to be a little careful though because remember, t is time in days, and this question is giving us the time in weeks. And since there seven days in a week, seven weeks is equal to seven times seven or 49 days and therefore, we use 49 for the time t to determine the mass left after seven weeks, so now we need to find p of 49. So we substitute 49 for t, which gives us 48 times e raised to the power of 1/4 natural log 1/2 times 49.

And we go back to the calculator one more time. And we are told to round to four decimal places or enter the exact value which should be this value here. Let’s go ahead and find the decimal proximation so we enter 48, again, second natural log brings up e raised to the power of 1/4, which is equivalent to 0.25, so I’ll enter 0.25, and then natural log 1/2, which is equivalent to 0.5, close parenthesis, and then times 49, enter.

Notice how there’s a five in the fifth decimal place, and therefore, we found up to 0.0099. So now we know after seven weeks, approximately 0.0099 milligrams of the palladium-100 is left. I hope you found this helpful..

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