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