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welcome to our final video on exponential and logarithms. In this video, we're going to look at the properties of logarithms and how we can use them to solve and for the equipped in the equations where we actually have logarithms. Showing up in our physics class is the most common place that you're going to run into a log rhythms in physics 101 is when we talk about sound in particular when we talk about decibels Thean Quay, Asian for calculating decibels Looks like this. We say that beta is equal to 10 decibels times the natural or not the natural log, but log based 10 of some sound intensity I divided by I not. Okay, so this is an interesting equation. First of all, because we have this log based 10 here instead of a natural log, as we see in a lot of other places in physics that we end up with E um and this is because this is actually a defined. This is equation that was simply created to say, Hey, let's create a scale for sound that doesn't use very large numbers, because intensity, for example, I not here as measured in in S I units is going to be one times 10 to the negative 12 watts per meter squared. Don't worry about the units too much and what they are, but yeah, one times 10 to the negative 12 is not a comfortable number to report. And so what they did was they created a log scale that would allow them to give numbers that they're a little more comfortable things that are like in the tens and the hundreds. Um and thus we use the base 10 scale here. Eso if I were to say, Hey, what's the particular intensity I when I have ah reading of, say, 100 decibels on this side? Well, then I can say 100 decibels is equal to 10 decibels times log base, 10 of I over Aina and then similar to the exponential Zai can use a lot of the same steps that I did before by 100 divided by 10. That's going to be 10 is equal to log base 10 of I over I not. And then the inverse operation of log based 10 is tend to that power, so we'll have 10 to the 10 is equal to tend to the log base 10 of I over I not which will end up just being I over Aina And so we have I is equal to I not times 10 to the 10 which this looks like a huge number But remember, we just had China is 10 to the negative 12 which is actually much smaller than 10 to the 10 is big, so we actually end up with one times 10 to the negative tube watts per meter squared. Okay, so that's one way to handle this. Another way that we can handle it is actually by using property of logs to separate out the eye and the eye. Not so Let me show you how that works. So again we come back to this equation. Beta is equal to 10 decibels times the log based hand I over. I not remember that I can rewrite this thing as 10 decibels Times log based 10 of I minus log based 10 of China. And this can actually be a really helpful thing to look at because remember, I not is this extremely small number and in fact it's a constant, which means this is just another constant. It's a negative constant. So when we put this in here, we get in the media answer and we could solve Justus quickly. We'd have beta plus 10 decibels. Log based 10. I not divide the whole thing by 10 vegetables so that we end up with log based 10 times I on a side by itself. And then we can do the same things before we take 10 to this power, which is a little intimidating. But it works the same as anything. And then we have tended. This will end up just being I and you'll get the exact same answer as we had before s Oh, this is an illustration of how the using the inverse property of those two functions can be a little bit cleaner. Um, the occasionally because I also has things that that we used to calculate it. So if we will break it down in those pieces, it might actually end up being more useful to us to use these properties of logs to separate it out into individual terms.

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