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In this video we practice using the properties of logarithms and we do the following five problems.
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So try these two problems. Write each expression as a sum or difference of logs.
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And we're going to assume all variables represent positive numbers.
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So what we have to start with with number one is, well this is a quotient.
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So I'm going to begin by doing subtraction.
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So that gives me log base three of the numerator which is 2x minus the log base three of the numerator which is square root of five.
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Now let's write square root of five as five to the one half because it's better to take anything in radical form and use exponential form.
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So that's what we have so far.
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Okay, second part.
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I have the log of 2x. Notice that's a product.
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So now I have to go ahead and rewrite the log of 2x base three as the sum of the logs of each of those that are in the product.
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So I'm going to have to move this down, kind of get it out of the wafer now. So I have room to do that.
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So that gives me the log base three of two plus the log base three of x.
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So what I've done, I have taken this part here and written it as log of two base three plus log of x base three. Now that is using the product property.
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Let's go into the next part minus the log of five to the one half base three.
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I could take that exponent, put it out in front. That'll give me a minus one half log base three of five.
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And that is as far as we're able to go.
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Now if you didn't get that, write this problem again on a piece of paper.
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Don't look at this and see if you can get the answer on your own.
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All right, let's try the second one.
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So I've got the log base two of five m cubed.
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All right, I do have something cubed, but the overriding thing is first I have a multiplication five is not to the third power.
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That's really five to the first power. So it's five times m cubed.
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So the first thing I need to do is use the product property.
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So I have the log base two of five plus the log base two of m cubed.
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All right, can't do anything with this log of five base two.
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So I'm going to leave that as log of five base two.
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But now I could take this exponent from the m and put it out in front.
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And that gives me three log of m base two.
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And that's the second one.
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All right, you ready for some more?
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All right, let's try this expression. This is going to be a little bit more involved.
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So try it on your own. See how you do?
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I put in the video on pause.
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Okay, let's tackle it.
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So the first thing is I have a quotient.
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So again, I'm going to start off by writing it as a difference.
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The log of the numerator minus the log of the bottom.
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So I have the log base five of 25 square roots of seven
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and minus the log of five m cubed.
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The trick and math is always breaking things up into smaller little pieces.
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Okay, now let's look at this first log.
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I have a product. It's 25 times the square root of seven n.
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So you want to break that up into a sum, right?
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So we have the log base five of 25 plus the log base five of square root of seven n.
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Now remember, that's the same thing as seven n to the one half power.
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Okay, the trick is remembering.
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Anytime you're working with radical signs, you could write the problem with exponents.
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And let's see, can I do anything over here?
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The log base five of m cubed.
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Well, at this point I could bring the three out to the front.
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I could use my product property.
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So that'll be three minus three log base five of m.
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All right, can't do anything more with this.
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What about this stuff over here? Can we simplify this anymore?
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You should know this one. What's the log of 25 base five?
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Five to the what power actually equals 25.
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Do you get it? It's two.
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Remember, logs are just exponents. That's why it's a small little number.
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Oops, now we got to deal with this. What do we do with this?
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Seven n to the one half.
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Like to put the one half out in front.
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And then I have log base five of seven n.
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Now you have to put a parentheses around the seven n.
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Actually, you don't have to, but it's a good idea.
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I better say it that way.
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All righty.
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It'll I can't simplify too many for this.
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So I have two. Now what am I going to do with this?
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Ah, I got a product seven times n.
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So now I could break that up.
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So I have one half.
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And I could rate that as a sum.
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The log base five is seven plus the log base five of n minus this three log base five.
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And the last thing you might want to do is distribute the one half.
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So it just looks like the sum and difference in numbers,
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which is what the directions had to do.
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So I've got two plus one half, log of seven base five, plus one half, log of n base five, minus three.
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And the log of n base five.
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Okay, well, now that was a difficult problem.
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Just practicing using the properties of logs.
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All right, you're going to go the other way around now.
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You're going to write each of these as a single logarithm.
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And you should actually be able to simplify five after you do that.
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So go ahead and try these by putting the video on pause.
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Okay, let's do the first one.
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So we've got...
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Well, I could take these exponents.
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I mean, I'm sorry.
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The numbers in front of the logs and put them as exponents.
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So this will be the log of x cubed.
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I mean, a mistake or I forgot to put the base.
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So let's say the base was four on both of these.
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Sorry.
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Minus, and I've got the log base four of five to the one half.
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And remember what five to the one half is.
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That's just the square root of five.
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So let's see, move this down a little.
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All right, now I have a subtraction of logs.
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So you can write that as a question.
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So I've got the log base four.
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The numerator will be x cubed.
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And the denominator will be five to the one half or square root of five.
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Okay, you leave the answer like that.
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Or if you want, you could put parentheses around it.
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So it does get confused with this base of four.
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All right, let's go on to this next one.
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Okay.
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Well, I have a minus. This thing's going to go in the denominator eventually.
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And this one is a plus.
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So it's going to go in some numerator, right?
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The positives will go in the numerator.
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But I first have to deal with this two out in front of the log.
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So that's going to be the exponent on six.
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So let's begin with that.
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This will give me a log base three of six squared minus the log base three of eight plus the log base three of two.
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And six squared is 36.
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So I have the log base three of 36 minus the log base three of eight plus the log base three of two.
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Right now, I'm subtracting here.
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So that is the log base three of 36 over eight.
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So I can begin with that.
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36 over eight.
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Some people might do this in one step.
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I'll do that next.
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But for now, I'll just go left to right.
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That's what we have so far.
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Now, we could reduce that now.
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I'm going to wait and do that at the next step.
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So that I have log base three.
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Now, I have a sum.
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So I multiply these two.
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So I basically have 36 over eight times two,
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which I'm going to write as times two over one.
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I'm just going to do a little canceling here.
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I end up with 36 over four, which is nine.
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Right?
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So I get...
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Oops.
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A little bit too far here.
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Log base three of nine.
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Huh!
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And what's that?
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Three to the what power is nine?
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Two.
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I like that problem.
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It's kind of cool.
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Okay, so let's look at it again.
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This was the beginning.
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And we went all the way down to two.
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Now, this step right here could have been skipped.
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You could have gone directly to this step.
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Anything with the plus signs in front.
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Those go in the numerator.
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So the 36 and the 2 go in the numerator.
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And the 8.
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That was minus.
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That would end up in the denominator.
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So there is one step you could have written a little bit differently than me.
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Okay, that's it.