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Function Notation with Logs and Exponentials - Problem 2 4,752 views

Teacher/Instructor Carl Horowitz
Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

So we are now going to look at a applied exponential equation in function notation. So what we have is a set up where the amount of radioactive material decays such that the amount remaining after time t is given by this equation in function notation, and our units of measure are grams for the amount and I didn’t specify t but it’s going to be in years.

So the first question is find the initial amount of material, and the trick for this is finding what the time is in the initial amount. Initial amount means starting and its remaining after time t so initial amount is after time, no time is disappeared so time is just equal to 0.

Anytime we are dealing with function notation we just plug in our variable so this is just turns into f(0), 100 times 2 to the 0. Anything to the 0 power is just 1 so what this leaves us with is 100 grams as our unit of measure. So our initial amount is 100 grams.

The second part is asking how much is going to be left after 4 years? It’s basically the same exact premise except instead of dealing with t is equal to 0 we now know that t is equal to 4. Plug in f(4) is equal to 100 times 2 to the -4 and then solve. 2 to the fourth is 16, the negative exponents puts in the denominator so what we have here is 100 over 16, simplify this out divide each top amount by 4 leaving us with 25 over 4 units of measure once again units grams.

Last problem is a little bit different, so instead of talking about how much is going to be left after some time, we are asked how much time is going to go by to get to a certain amount. So instead of plugging in for t where we are actually solving for t. And we are told that the amount left is 50 and f(t) is actually representing the amount left.

So for this problem we said 50 equal to our equation and then solve for t. And this is a little bit different than anything we’ve done, but with one little step it becomes very similar to solving any other exponential equation and that is just get your variable, get your exponential term by itself. If we divide by 100 what we end up with over here is 1/2 is equal to 2 to the -t. Now you could rewrite get your base is the same but hopefully you are getting into the point where you can recognize what these powers are.

2 to the first is just 2, the negative we’ve put in the bottom so this is just going to be t is equal to 1. 2 to the negative of first is 1/2. So different ways of solving for different things depending what you are asked for using a exponential equation in function notation.

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