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0698The first thing: what kind of formula are we working with here?0707Well, we have compounding, but not continuous compounding; so we go and look that up.0710It is principal, times 1 the rate, divided by the number of times that compounding occurs,0714raised to the number of times that the compounding occurs, times the amount of time elapsed in years.0721So, our principal investment here is P = 4700; and we know that, at time 5, at t = 5, we have 5457 dollars and 57 cents.0727So, 57.57 is equal to..was our principal amount? 0743That is one of the things we don't know yet--we don't know what our rate is.0754And that is why we are setting up at t = 5, instead of just hopping immediately to the 10 years question:0758we need to figure out what our rate is first, so that is what we are figuring out now.07631 r/n; our n is quarterly, so that is an n of 4, because it happens four times in a year, in each of the four quarters of the year.0767So, 1 r/4, raised to the 4 times t--do we know what t is in this case?

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--you probably wonder.0285Why can you do this--how can you get away with using Pe can morph into other forms.0297For example, let's look specifically at a possible half-life formula.0303We might have P times 1/2 to the t/5; we can see this as P times 1/2 to the rate of 1/5 times t.0307That is what we have there: some principal starting amount, times 1/2 to the 1/5 times t.0318So, for every, say, 5 years, we have half of the amount there that we originally had.0323So, how can we get Pe form right here; we know what r is, so we swap that in for our r.0343And we get P times e to the -0.1386 times time.0350But we also know that -0.1386 is the same thing as -0.6931 times 1/5.0358So, if we want, we can break this apart into a -0.6931 part and a 1/5 times t part that we might as well put outside.0366We have e to the -0.6931, to the 1/5 times t, because by our rules from exponential properties, that is the same thing0374as just having the 1/5 and the -0.6931 together, which is the r that we originally started with.0381Now, it turns out that e to morph into something else.0396We can get it to morph into this original 1/2; and now, we have this 1/5 here,0402so it becomes just P times 1/2 to the t/5, which is what we originally started with as the half-life formula.0406So, by this careful choice of r--and notice, the r here is equal to 1/5; the r here is equal to the very different 0.1386;0414we get totally different r's here; but by choosing r carefully, if we have enough information from the problem0426(sometimes you will; sometimes you won't; you will have to know that special formula)--sometimes,0433you will be able to get enough information from the formula, and you will be able to figure out what r is.0438So, you can have forgotten the special formula--you can forget the special formula occasionally, when you are lucky.0441And you would be able to just use Pe, we morph it into something that works the same as the other formula.0448Now, of course, you do have to figure out the appropriate r from the problem.0454You are just saying it--you have to be able to get what that r is.0457And remember: it is the r for Pe, which may be (and probably is) going to be totally different0460than the r for the special formula that we would use for that kind of problem.0465But if you can figure out what the r is from the problem, you can end up using Pe instead.0469Once again, we will talk about a specific use of this in Example 2, where we will show how you can actually use this if you end up forgetting the formula.0473Now, I want you to know that the above isn't precisely true.0480e isn't precisely 1/2; it is actually .500023, which is really, really, really close to 1/2; but it is not exactly 1/2.0484But it is a really close approximation, and it is normally going to do fine for most problems.0496It is such a close approximation that it will normally end up working.0500And if you need even more accuracy, you could have ended up figuring out what r is, just to more decimal places.0504And you could have used this more accurate value for r.0509Applications of logarithmic functions: logarithms have the ability to capture the information of a wide variety of inputs in a relatively small range of outputs.0513Consider the common logarithm, base 10: if we have log(x) equaling y, log(x) going to y--0522over here we have our input, which is the x, and our output, which is the value y--that is what is coming out of log(x).0528x can vary anywhere from 1 to 10 billion; and our output will only vary between 0 and 10.0537That is really, really tiny variance in our output, but massive variance in our input.0545Why is this happening?

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