Hey, ladies and gentlemen, this is mrs. Beckmann here
and today we're going to be going over unit ten video three, so we're gonna be talking about
properties of logarithms
so
let's just go ahead and let's just start at some of those properties because we've briefly talked about logarithm and now we're going to kind of
put it into example. So one of the things that you need to remember is on that product of powers property.
so for example,
if you have three to the second power times 3 to the fourth power, to simplify that out
you would actually add the exponents, and
that would give you 3 to the sixth power.
Okay, so logarithms are gonna be following that same rule.
Okay, and also if you're thinking back to that quotients property, okay
because it looks like I don't have a slide for that, when you're thinking about the quotients property as well looking at that, second one
if I have three to the fourth divided by three to the second power, then I'm gonna be subtracting those exponents.
So I have 3 to the 4 minus 2, which is 3 squared. So those same properties are gonna follow through for logarithms.
So a logarithm as a product of a product is the sum of the logarithm of its factors.
So for example if I have log base B of M times N, that's the same as log base M plus
log base N.
So if we have something log base 3 of 2 times 4, I can split that up
to be log base 3 of 2, plus log base 3 of 4.
Okay,
so let's go ahead and let's take a look at this first problem.
So here it says use log base 4 of 2 is equal to 0.5 and log base 4 3 is equal to
0.79248 to approximate log base 4 of 6.
Well if you think of 6, 6 is the same as 2 times 3.
So what I can do is I can write out log base 4 as 2 times 3.
And I can use those log properties to split it up. So log base 4 of 2
Plus log base 4 of 3. Well, I know what log base of 2 is, and that's about point 5.
And log base 4 of 3 about 0.79248
So now all I have to do is add those decimals together and that is going to give me 1.29248.
So that's going to be my answer for number one.
Now, let's take a look at number two.
So for number two, use log base two of three is equal to 1.5850
to approximate the log value of 48.
So let's take a look at 48. So 48, I have 2 and 24.
I want to try and get this into twos and threes if I can. So two and twelve, two and six, and
six and two.
Okay. So now what I'm going to do is I basically have a log. So I have log base two of 48,
Which is really the same as two times two times two times two. Oops. This should be a three
And so I have four twos and a three.
okay, so now that's really the same as log base 2 of
2 to the fourth power plus
Log base 2 of 3. Well, why is log base 2 of the fourth power relevant?
Well, first of all when I have an exponent on the outside,
Remember what we did on the previous note sheet, that you can drop that down.
So I have log base 2 of 2, plus log base 2 of 3, which I know is 1.5850
Well, when I have the same base and the same number, that's actually going to be equal to one.
So this is 4 and then this whole log right here is equal to one.
So I have 4 times one plus 1.5850
so when I add those together, I get 5.580
And that's my answer for number two.
Now, let's take a look at number three. So we want to solve for X.
Okay, so to solve for X,
I want to start by writing it as one log.
So I have log base four of X times X minus six is equal to two.
Now, I want to write it as one log whenever I'm doing an equation.
Okay, so I'm going to combine the separate logs into one log. Now I want to write it in exponential form.
So this right here is my base and this guy over here is my exponent. And that's going to be equal to
X squared minus 6X when I distribute that so I get 16 is equal to X squared minus 6X.
So then my next step is to get that 16 over to the other side.
So then I have x squared,
Minus 6x minus 16 is equal to 0.
so now I need to factor that. So when I factor it I need to think about what
Multiplies to a negative 16 and adds to a negative six.
So I'm going to be doing X plus 2 and X minus 8.
So now we're going to set each of those equal to 0, so X minus 8 is equal to 0 and
X plus 2 is equal to 0.
So here I get that X is equal to negative 2, and here I get that X is equal to 8.
Well what we talked about logs, we talked about the fact that we cannot take the log of a negative number.
So since we have log base 4 of X here,
We have to throw up that negative number because we can't have a negative with our log.
So that means that our only answer is X is equal to 8.
Now let's take a look at number 4.
So the first thing we're gonna do is combine it to one log.
So I get log base 4 of  X times X minus 12 is equal to 2. So that there is my base.
So I have 4, and then I'm raising it to the second... oops
I wrote 4, it should be 8. Log base 8.
So I'm taking 8 and raising it to the second power and that's equal to when I distribute X squared minus 12X.
Well 8 squared is
Going to give me 64, so I need to bring that over to the other side. So I get X squared minus 12X
Minus 64 is equal to zero.
So now I need to factor this, so I'm going to set up my two factors.
So I'm gonna put an X in one, a plus in one, and a minus in the other.
So I want to multiply to a negative 64,
And add to 12, and that's going to be a negative 16 and a positive 4 is equal to 0.
So then I'm going to set each factor equal to 0.
So here, I get that X is equal to a negative 4. Well, we can't take the log of a negative number,
So I'm going to throw that out and here we get X is equal to 16.
So X is equal to 16 is our answer for number 4.
Now let's take a look at talking about the quotient of powers, and we talked about this briefly on the front
Ok, so when you're dividing things of like bases, you are going to subtract the exponents.
So 3 to the fourth divided by 3 squared is going to be 4 minus 2.
So that's going to give you 3 squared.
And so since logarithms are the same as exponents, you're going to use that same property.
So I have log base B of M divided by N, that's the same as log base M minus log base N.
So let's go ahead and let's take a look at that as an example for this next one.
Okay. So what we're going to do is we're going to simplify this, so I'm going to have log base 3 of 4.
So what is on top is always first, and then what's on the bottom or the denominator of your fraction,
That's what's going to come after the subtraction symbol. So this is gonna become,
log base 3 of 4 minus log base 3 of 2.
Ok, so let's go ahead and let's take a look at this.
So now we're going to go ahead and use this. So we have
Log base 3 of 5 is equal to 1.4650
and log base three of 20 is equal to 2.7268
So we want to use that to approximate the value of log four.
So I want to think about how I can use five and 20 to get 4.
Well 20 divided by 5 is equal to 4.
So I'm going to have log base 3 of 20 divided by 5.
So when I go to split that into two different logs, I'm going to have log base 3 of 20
Minus log base 3 of 5. So now that I got it to two separate logs, I can go ahead and subtract that.
So I'm going to have 2.2768, because that's what log base 3 of 20 is equal to,
minus 1.64650 and when I subtract those two, I get 1.2618.
Now we talked about this a little bit earlier, but when you're doing a product of powers, you're going to multiply their exponents.
So remember that if I have something that is three to the fifth power raised to the second power,
I'm going to multiply those exponents to get three to the tenth power.
So this actually can be used for logarithms as well. So when you're doing logarithms, if you have an exponent on the outside,
remember that that's going to drop to the outside of the log.
So if something in the log is being raised to an exponent that's going to drop to the outside of the log.
So that's what's gonna happen here. We have this two.
So I need to take this two and drop it to the outside of that log.
So that 2 is gonna come on the outside. So I have two times log base three of five.
And then that right there would be the simplified version.
So let's go ahead and let's use this property to finish these last couple of problems.
So here it says use the fact that log base four of six is equal to 1.2925.
So I want to figure out how I can use properties of logs,
so I need to figure out how I can take 36 and
Somehow get it to be a six, and I can use exponents to do that.
Well, 36 to the second power is...
sorry, 6 to the second power is 36. So I'm going to have log base four of six to the second power.
So now I can take this exponent here, and drop that to the outside. So I'm going to have two times
log base four of six.
Well, I know that log base four of six is 1.2925.
So I have two times 1.2925, and when I multiply those together I get 2.585.
So that's going to be my answer for number six.
Now, let's take a look at number seven. It wants us to solve this log.
So the first thing I want to do is I want to combine it so that everything is one log.
So first of all this 3 on the outside, I need to use to bump up to that exponent.
I'm trying to get it down to one log on that side. So that's gonna become X to the third power minus
Log base 5 of 4 is equal to log base 5 of 16.
So now I need to write these two as one log.
so I get log base 5 of X to the 3rd divided by 4, is equal to log base 5 of 16
so now I have it as one log on either side and
Since they're both the same log, I can set them equal to each other.
So X to the 3rd divided by 4 is equal to 16.
So to solve this, I'm gonna start by multiplying both sides by 16.
So when I do that, I get X to the third... oops, not by 16. By 4, silly me.
I'm gonna multiply both sides by 4. So 16 times 4 is 64.
So now I need to take the cube root of both sides,
Which if you remember is the same as taking 64 and raising it to the 1/3 power.
And when you do that you get 4. so here, my answer is that X is equal to 4 for number 7.
Now, let's take a look at number 8. So for number 8, we want to start by bumping this exponent up.
So we're going to bump that exponent up. So I'm gonna get log base 2 of X to the 4th power,
Minus log base 2 of 5 is equal to log base 2 of 125.
So now I want to combine this to be one log.
so I have log base 2 of X divided by 4... I mean X to the fourth divided by 5 is
equal to log base 2 of 125.
So since they're all log base 2, now I can take it and set them equal to each other.
So I have X to the fourth divided by five is equal to 125.
So I'm gonna start by multiplying both sides by five.
So I get X to the fourth is equal to 125 times five, So that's going to be equal to 625.
So to figure out what X is equal to, I need to take the fourth root of both sides.
So remember, that's the same as taking 625,
And raising it to the 1/4 power. And when I do that I end up getting X is equal to 5.
So that's my answer for number 8.
Now, let's take a look at number 9. Ok. So again, I'm going to start by raising that as my exponent.
So I have log base 7 of
X squared, is equal to log base 7 of
27, plus log base 7 of 3.
So since these two logs are being added, we can write them as one log being multiplied.
So I have log base seven of X squared is equal to log base seven of
27 times three. So that's going to give me log base seven of X squared is equal to log base seven of 8.
So now we have both of these as the same log base, so I can take those two and set them equal to each other.
So I get X squared is equal to 81. So I'm going to take the square root of both sides.
So I get X is equal to plus or minus 9. Well, we can't have a negative log,
so we end up getting that X is equal to 9.
That's our solution for number 9 and that concludes your note video for today. Thanks for listening!
