WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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We are going to continue our discussion of logarithms by talking about special logarithms.
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And the first one is common logarithms.
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First of all, what are common logarithms?
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**Common logarithms** are logarithms to the base 10; and these are used very frequently in many real-world applications.
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For example, the scale that is used to measure the magnitude of earthquakes is base 10 scale.
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And the base 10 is not written: logs in base 10 are just written as follows: log(x).
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So, if I were discussing log < font size="-6" > 10 < /font > (7), I could instead just write that as log(7).
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And it is assumed that, if nothing is written here, then this is base 10--
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in the same way that, if you took a square root of 4, we know that this really means this;
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but we don't write the number (because it is so commonly used)--we don't write the number 2.
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If we are talking about some other root (like the third root--the cube root), then we would write it.
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The same idea here: if you don't see anything here, you can assume that it is a base 10 log.
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Now, one application that we are going to have with this is to use logarithms to help us solve exponential equations.
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In previous lessons, we used exponential expressions to help us solve logarithmic equations.
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When we had a logarithmic equation with a log on only one side, we converted that to the exponential form.
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Well, it works the other way as well: there are times when taking an exponential equation
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and taking the log of the equation can actually help you to solve it; so let's talk about that right now.
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If we are working with an exponential equation, the previous techniques we used involved having the bases be the same.
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For example, given something such as an exponential equation 3^x + 1 = 9^4x,
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we were only able to solve this if we had the same base.
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And sometimes I could convert it to the same base if I didn't have the same base, because 9 is equal to 3²;
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therefore, I would end up with 3^x + 1 = 3^8x, and then from there, you can solve,
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because if the bases are equal, then the exponents must be equal.
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However, getting a little bit more advanced: you are going to run into situations like this: 2^3x + 4 = 5.
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These are not the same base, nor can I easily convert them to the same base.
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In this case, if both sides cannot be written as powers of the same base
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with an exponential equation, solve the equation by taking the common logarithm of each side.
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The common logarithm means the base 10 log, which I am just going to write without the 10 here, as is standard.
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So, log (and that is actually base 10) of 2^3x + 4 = log(5).
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Once I have it in this form, then I can solve: you treat this as you would any other equation, which is just by isolating x.
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Now, let's first rewrite this, using the power property, so that we get rid of this exponent.
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Let's just go ahead and write this as a coefficient: (3x + 4)log(2) = log(5).
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I have my variable in here; I want to isolate the x, so I am going to divide both sides by log(2).
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Keep in mind that these are just numbers--there are no variables in here.
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log(2) has a specific value; log(5) has a specific value; I haven't found that value, but having my answer in this form is perfectly valid.
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If you look on your calculator, there is a log button; and that log button, if it just says 'log,' is for the base 10 log,
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although on some calculators you can specify other bases.
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This is something you can easily find the value of.
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So, by doing this, now I have the x much more freed up: I am going to subtract 4 from both sides.
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And now, all I have to do is divide each side by 3.
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And instead of writing this as a complex fraction, log(5)/log(2)/3, I can simply remember that this is the same as this.
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So, if I said log(5)/log(2) times the reciprocal (which would be times 1/3)...I want to get rid of that complex fraction,
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so I am going to write it like this, in a more simplified-looking form.
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OK, therefore, I started out where I had an exponential equation where there were different bases.
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And I couldn't convert them to the same base very easily at all.
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So, I went ahead and just took the common log of both sides, then used the power property in reverse to get this over here as a coefficient.
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Next, I divided both sides by log(2), because that is really just a number.
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And I isolated the x on the left; and everything I have here, I either have a value for, or I can find a value for.
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The same techniques that we used for solving exponential equations can also be used to solve inequalities--
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exponential inequalities where we can't get the same base very easily at all.
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Recall that, when you are multiplying or dividing both sides of an inequality, you need to make sure that you are not working with a negative.
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And usually, it is obvious: in past lessons, we have known if we are dividing by -3 or multiplying by -3;
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we need to flip the inequality sign--it is obvious.
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But you have to be careful with a logarithmic expression, if you are taking the log of some number,
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that you check and make sure that you are not ending up with a negative,
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inadvertently, and then having the inequality symbol be in the wrong direction.
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All right, so let's look at 4^3x - 5 > 3^2x - 6.
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As we did with equations that were exponential equations with different bases, we can do the same technique with this inequality,
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which is to go ahead and take the common log of both sides.
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So, this new inequality is still valid, because I did the same thing to both sides of the inequality--
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the same as if I had added a number to both sides or multiplied both sides by a number.
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This is still a valid inequality: the relationship between the left and right sides is still holding up.
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Next, I am going to use the power property to bring this out in front, because my goal is to isolate x.
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And I can't isolate x when it is up there as a power.
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3x - 5 times log(4) is greater than...bring this out in front...2x - 6 times log(3).
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All right, next I am going to divide both sides by log(4).
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Actually, we need to take one more step: we can't just go ahead and do that.
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What I need to do is separate this out; I need to use the distributive property to split this out,
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because what I really have here is 3xlog(4) (that is 3x times log(4)--we can look at this like this--it makes it much easier), minus 5log(4).
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OK, and this is greater than 2xlog(3) - 6log(3).
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Now that I have this, and it is split apart, I can go ahead and look at the expressions that do not contain variables.
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This does not contain a variable, and neither does that; I want those all on the right, and my variable-containing expressions over on the left.
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I am going to start out by adding 5log(4) to both sides.
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My next step is going to be to subtract 2xlog(3) from both sides.
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All right, what I have now is expressions containing variables on the left, and those with constants only are on the right.
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And again, this is really just a number.
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Let's go back up here and continue on.
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And my next step is going to be to factor out x from this left side of the inequality.
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And this is going to give me an x here, times 3log(4), minus 2log(3), is greater than -6log(3) + 5log(4).
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So, looking at what I did going from here to here, I just factored out an x.
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I pulled that x out of here and here, leaving me with this difference: 3log(4) - 2log(3).
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Now that I have done this, I can isolate the x at last.
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I just keep x on this side, and I divide (I am going to rewrite this with 5log(4) in the front
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and the negative second, because it is more standard) by 3log(4) - 2log(3).
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OK, so in order to solve this exponential inequality that had different bases,
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I proceeded by first taking the common log of both sides,
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then isolating the x (which was a little bit complicated).
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I used the power property, and then I used the distributive property
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to separate out this 3x and the 5, since this contains a variable and this doesn't; and the same here.
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Then, I went ahead and factored out an x, and then just divided both sides by this expression.
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So, I moved the constants to the right, and the variable-containing expressions to the left, and then divided.
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Now, as I mentioned, you have to make sure that you don't end up dividing by something negative inadvertently.
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So, at the division step (that was this), let's go ahead and take a look at this.
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I know that I am OK, because the log of a number greater than 1 is positive.
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So, if I take the common log of a number greater than 1, it is positive.
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I know this is positive, and I know that the log of 4 is going to be bigger than the log of 3.
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And I know that 3 is bigger than 2; so this is going to be larger than this, so I know I am working with something positive;
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and I don't need to flip the inequality symbol--but that is important to check.
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The last new concept we are covering in this lesson is going to be change of base.
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There are times when you have a log given in one base, and you want to change it to another base.
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Frequently, it is base 10 that you are wanting to change it to, but not always; it could be a different base.
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And so, this change of base formula allows us to write a given logarithmic expression in a different base.
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And it allows us to evaluate logarithmic expressions in any base by rewriting it using common logarithms.
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And as I said, frequently what you want to change the base to is a base that is log base 10.
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What this right here is, is the original expression.
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This is actually the original expression; and what we do is divide that original expression by a log to the original base.
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And what we take the log of is a value that is equal to the new base.
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So, I take just the original expression and divide it by a log to that base;
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but for this number, I use the new base that I want.
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Illustrating this: let's say I had log₆(8), and I want to write it as log < font size="-6" > 10 < /font > .
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So, I want to write this as log < font size="-6" > 10 < /font > ; I don't need to write the 10 here.
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This with a new base would be equal to log₆(8), the original expression,
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divided by a log with the original base (base 6), but for the x value right here, I am going to use 10; that 10 is implied.
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So, if I needed to, for some reason, change this to base 10, this is equivalent to log < font size="-6" > 10 < /font > (8).
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You just need to learn this formula and follow it, and know that it is the original expression,
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divided by a log of the original base; and you are taking the log of the value equal to the base you are looking for.
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OK, in Example 1, we are looking at an exponential equation where the bases are not the same, and I can't easily get the bases the same.
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So, I am going to use common logs to solve this.
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I am going to start out by taking the common log of each side.
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And I am allowed to do this, as long as I do the same thing to both sides.
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Once I have it in this form, I want to isolate x; so I need to use the power property to get (2x - 3)log(3) = log(19).
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To isolate x, I divide both sides by log(x), because my x is here; I want to split that away.
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Next, I am going to add 3 to both sides, so that the 3 ends up on the right.
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And then finally, I am going to divide both sides by 2.
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Let's come up here in the second column: 2x = log(19)/log(3), divided by (2 + 3).
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And to make this look at little better, we can rewrite this as...actually, the x is now isolated, because we divided by 2;
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so x equals 1/2, because if I take log(19)/log(3), divided by 2, that would be the same as multiplying by the reciprocal, 1/2.
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So, I was able to solve this, even though this exponential expression had different bases,
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by taking the common log of each side, and then isolating x.
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Write using common logarithms: log₇(22).
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I need to use my change of base formula, which is log < font size="-6" > a < /font > (x) = log < font size="-6" > b < /font > (x), divided by log < font size="-6" > b < /font > (a).
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And here, a is the new base I want, and b equals the original base--the base that I already have.
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I want to find an expression that is equivalent to an expression with the original base.
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OK, log < font size="-6" > 10 < /font > (22): this is what I want, and I need to find an expression that is equivalent to that.
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And this would be just the original, log₇(22), divided by log with the original base;
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and then, for a, I am going to use this base that I want.
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So now, I have written this using common logarithms.
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OK, again, I have an exponential equation in which there is no common base, and I can't easily get a common base.
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So, I am going to take the common log of both sides to help me solve this.
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Now, I need to isolate x; so I need to get the x out of the exponent--I am going to do that by using the power property.
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And this becomes the coefficient (2x - 6)log(5) =...I am going to bring this out in front...(4x + 3)log(7).
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Now that I have done this, I am going to write it like this to make it clear what needs to be done.
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And we need to use the distributive property, because I want to split away these terms that have x in them.
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I am going to multiply 2x times log(5), and then I am going to multiply -6 times log(5), 4x times log(7),
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and remember, this is base 10 that we are talking about; and 3 times log < font size="-6" > 10 < /font > (7).
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Next, I am going to get all of the variables on the left, and expressions containing constants only on the right.
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And I am considering the log of 5 to be a constant, because we can find a specific value of that.
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So, I am going to add 6log(5) to both sides.
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The next thing I need to do is subtract 4log(7) from both sides, so I can get this on the left, because it does contain a variable.
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So, 4x - 4x log(7); now I have my constants on the right and variables on the left.
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The next thing to do: I want to isolate x, so I can factor out an x, because there is an x factor here and one here.
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I pull that out; I leave behind 2log(5); I pull out the x; I leave 4log(7).
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And I am almost there; all I need to do now is divide both sides by this expression, and I get x = 3log(7), plus 6...
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and this is log...that is a 5, not an exponent of 5...all divided by 2log(5) - 4log(7).
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Again, the technique is to take the common log of both sides, isolate x by first using the power property,
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then the distributive property, adding and subtracting as needed to bunch all of the expressions and terms
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containing x's on the left, constants on the right; factor out x, and divide both sides by this expression.
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And now, I have x isolated, so I have solved for x.
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This time, we are going to solve an inequality; and it is an inequality involving exponential expressions
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without a common base, and where I can't easily find a common base.
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So again, as I did with the exponential equation, I am going to take the common log of each of these.
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My next step is going to be to isolate the x.
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All right, so in order to do that, I need to get the x out of being an exponent, so I can use that power property.
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4x - 5 times log(3) is less than...this becomes...(-3x + 5)log(6).
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Further separating out the x, I need to use the distributive property: this is going to give me 4x...
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this is equivalent to this...so 4xlog(3) - 5log(3) is less than -3xlog(6) + 5log(6).
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As usual, we are going to move the variables to the left and constants to the right.
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So, I am going to add 5log(3) to both sides; and now I have this x on the right, so I need to move this term to the left by adding 3xlog(6) to both sides.
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I look on the left; I am still trying to isolate the x, and I see that these two terms have a common factor of x,
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so I factor that out to leave behind 4log(3) + 3log(6) < 5log(6) + 5log(3).
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Now, I am going to divide both sides by this expression; and I will have x isolated.
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Now, I am done; but I just need to double-check about this step, because, whenever you multiply or divide
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both sides of an inequality, you need to make sure that you are not multiplying or dividing by a negative number,
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because if you are, you need to reverse the inequality symbol.
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So, I am looking right here; and I divided both sides by 4log(3) + 3log(6).
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However, I know that I am OK, because the common log of a number greater than 1 is positive.
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So, this is positive; and I am adding that to something else positive; so I know that I am OK--that that comes out to be the log of a positive number.
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That concludes this lesson about common logarithms on Educator.com.
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