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 exploring some properties of logarithms.
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The first one is the **product property**: if you have a log < font size="-6" > b < /font > (mn), that is equal to the log < font size="-6" > b < /font > (m) + log < font size="-6" > b < /font > (n).
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And this might look familiar to you, or this general concept, from working with exponents.
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Recall that, when you are multiplying two exponents with the same base, you add the exponents.
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Since logs are a type of exponent, it stands to reason that there would be similarities.
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So, you are going to see some similarities between the properties we are going to cover today and properties from working with exponents.
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For example, log₉(10) = log₉...and I could break this out into factors, 5 times 2...
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equals the sum of the log of this factor, plus this factor...so the sum of log₉(5) and log₉(2).
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Or I could move, instead of from left to right, from right to left.
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And instead of breaking it apart into 2 separate logs, I may have to combine it into a single log.
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And combining into a single log is especially helpful when we move on to solving more complex logarithmic equations
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than the ones we have seen in previous lessons.
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For example, given log₆(5) + log₆(7), I may want to combine those.
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And this would be equivalent to log₆...since these have the same base;
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notice that they have to have the same base for this to work...of 5 times 7, which equals log₆(35).
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This, of course, applies when we are working with variables, as well--with logs involving variables, which is a lot of what we are going to be doing.
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So, log₈(x² - 36)...I could factor this into log₈(x - 6) (x + 6),
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and then write it as the sum of log₈(x - 6) + log₈(x + 6).
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And again, I can move from right to left: if I were given this, I could multiply these two and then combine this into a single logarithm.
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The **quotient property**: again, you are going to notice similarities from your work with exponents.
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If I have log < font size="-6" > b < /font > of the quotient m/n, this is equal to the difference between the log of the dividend and the log of the divisor.
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Given log₂(14/5), I could rewrite this as log₂(14) - log₂(5).
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Again, this works when we are talking about logarithms with variables (algebraic logarithms).
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log₃ of something such as (x² - 3)/y would be equal to log₃(x² - 3) - log₃y.
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So, the log of the dividend minus the log of the divisor--that is the quotient property.
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The third property we are going to cover is the **power property**.
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This one is a bit different that what you have probably seen before.
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But it states that log < font size="-6" > b < /font > of m to the power p equals p times log < font size="-6" > b < /font > (m).
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So, if you see what happened: we took this exponent up here and moved it to the front to become a coefficient.
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Again, you are going to find this property helpful when you are trying to simplify logarithmic equations,
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or when you need to work with equations to solve them, to be able to move back and forth between these two forms.
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For example, log₄(5³): I am going to take this 3 and move it to the front.
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It is going to become the coefficient: 3log₄(5).
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Or, another example using a variable: log₃(x⁶) =...I pull this out in front...6log₃(x).
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Again, these properties can be used to solve logarithmic equations.
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Earlier on, we talked about solving logarithmic equations where there was just a log on one side,
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and situations where we had one log on each side of the equation.
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Sometimes, however, you are given logarithmic equations where you have multiple logs on each side, sums and differences of logs...
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And if they have the same base, then you can actually combine those, so that you end up with one log on each side.
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And then, you can move on to use the techniques previously learned in order to solve those.
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The product, quotient, and power properties are some extra steps that you might have to take before applying previously-learned techniques.
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For example, given log₇(4x + 50) - 2log₇(3) = log₇(x + 1) + log₇(3),
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I see that these all have the same bases, so I want to use the property where, if I get a log on one side
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equal to the log of the other with the same base, then I can just say x = y.
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But I need to get this into that form; so let's see what we can do to use these properties to combine them.
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First, I see that I have a coefficient here; so I am going to use the power property.
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And with the power property, we talked about how log < font size="-6" > b < /font > (m^p) = plog < font size="-6" > b < /font > (m).
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So, I was moving that power to the front to yield this form.
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I can do the opposite, though: I can take this coefficient and move it up here and make it a power again.
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And I am going to do that: so, this first log I am just going to keep the same for now: log₇(4x + 50).
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Here, I am going to turn this into log₇(3), and I am going to put this back over here and raise the 3 to the second power.
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That is the only power I have; so I am just going to work with that for now.
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And I know that 3² is just going to be 9, so I can change that to 9 in this step: it is log₇(9).
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And then, I am going to apply the quotient property on the left, because I have the difference of two logs that have the same base.
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Using the quotient property, I can rewrite this as the log₇(4x + 50) divided by 9.
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Recall that the quotient property (this was the power property up here) told me that, if I have log < font size="-6" > b < /font > (m/n),
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that is going to equal log < font size="-6" > b < /font > (m) - log < font size="-6" > b < /font > (n).
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So here, I am moving from right to left: I am taking these and combining them--that is what I am doing right here.
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On the right, I am going to use the product property: the product property says that, if I have a log < font size="-6" > b < /font > of a product,
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that is equal to the sum of the logs of those factors (log < font size="-6" > b < /font > (m) + log < font size="-6" > b < /font > (n)).
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So, I am going to go ahead and combine these two into a single log, log₇(x + 1)(3).
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Now, I have gotten this in the form that I can actually solve, because I have a log on the left to base 7 and a single log in the right to base 7.
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And if these are equal, then this expression must equal this expression.
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Now, I just need to go ahead and solve: 4x + 50, divided by 9, equals...I am going to pull that 3 out in front: 3(x + 1).
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So, I am going to multiply both sides by 9 to get 4x + 50 = 27(x + 1); 4x + 50 = 27x + 27.
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I am going to go ahead and subtract a 4x from both sides to get 50 = 23x + 27.
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Subtract a 27 from both sides: I will get 23 = 23x, so x = 1.
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Now, I always need to check back and make sure that the solution is valid, because I want to make sure I don't end up taking the log of a negative number.
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Checking for validity right here: I have log₇(4x + 50).
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Let x equal 1, and check that: log₇(4(1) + 50); that is log₇(4 + 50), or 54.
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So, that is OK; let's look right here: the other one I have to check is log₇(x + 1), which is log₇(1 + 1), and that is just log₇(2).
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So, this one is valid; this is a valid solution.
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We used this technique, but we had to take a bunch of steps prior to that in order to combine these into a single log on each side of the equation.
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Before we go on to work some logarithmic equations, let's just talk about simplifying using the properties that we have already covered.
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And when I am asked to simplify a logarithmic expression, that means that I want to get rid of quotients.
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I don't want to be taking the log of any quotients, of any products, or of any factors raised to powers.
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The first thing I am going to do is work with getting rid of this fraction bar.
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And I am going to use the quotient property for that, because log₆(x⁴y⁵),
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divided by this polynomial expression down here, is going to be equal to...let's just say
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that this whole thing is going to be equal to the log base 6 of the dividend (the numerator), minus log base 6 of the divisor.
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So, log < font size="-6" > b < /font > (quotient) equals the log of the dividend, minus the log of the divisor.
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All right, so I got rid of that fraction bar; I don't have any more quotients; but what I do have is a product.
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I am going to apply the product property to just get rid of this right here.
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Instead, I am going to say, "OK, I know that this equals the sum of log base 6 of this factor, plus log base 6 of this factor."
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So, I have to remember that I have my negative sign out here; that is going to apply to each of these terms, once I split them apart.
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That is log₆(x - 3)³ + log₆(y + 2)⁶.
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All right, I still have one product (I don't have any quotients), so I am going to take care of this one next.
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This is going to be equal to log₆(x⁴) + log₆(y⁵), minus
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log₆(x - 3)³, plus log₆(y + 2)⁶.
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Next, I am going to apply the power property.
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And recall that the power property says that, if I have log < font size="-6" > b < /font > (m^p),
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that is going to be equal to p...this is going to turn into a coefficient...times log < font size="-6" > b < /font > (m).
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So, I am going to take these powers and move them out front; that is going to give me 4log₆(x⁴), plus
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5 (that is going to go to the front, also) times log₆(y), minus (3 is going to go out in front)
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3log₆(x - 3), plus...the 6 comes out in front...6log₆(y + 2).
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The last thing I want to do is move this negative sign and apply it to each term in here.
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And that is as far as I can go with simplifying: 4log₆...actually, this is gone now,
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because we have pushed that out in front...plus 5log₆y, minus 3log₆(x - 3).
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And then, the negative applies to this term, as well: minus 6log₆(y + 2).
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And this is simplified as far as I can simplify it.
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I look here, and I no longer have the logs of any quotients; I don't have the logs of any products;
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and I don't have the logs of any terms or expressions raised to powers.
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Now, instead of expanding out the expression, I am going to do the opposite.
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I am asked to write this as a single logarithm: so I am going to instead compress this into one logarithm--the opposite of what I did in the last question.
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Let's start with the power property: I know that, instead of 3log₄(x - 3),
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I can rewrite this as log₄(x - 3), and I am going to turn this into an exponent and write it that way.
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Plus log₄(x + 4): that is going to be raised to the fourth power.
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Minus log₄(x + 7): there is no coefficient there to turn into an exponent.
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There is here, though: log₄(x - 8); and this is going to be raised to the third power.
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Now, I have the difference here, which means that I can combine this by using the quotient property.
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I also, within these parentheses, have some addition; so I can use the product property to combine those.
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And I want to start inside the parentheses; so let's go ahead and do that and use the product property
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to combine this into log₄[(x - 3)³ (x + 4)⁴].
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So, it's log base 4 of this times this, minus...here, also, I have a sum; so I can combine that
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by taking the product of these two factors: log₄[(x + 7)(x - 8)³].
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So, this is this minus this; I can actually leave these brackets off at this point, because I have a single log.
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Now, we have to apply the quotient property, because we have a difference.
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This is log₄ and log₄; and this is going to become the numerator--it is going to become the dividend.
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(x - 3)³(x + 4)⁴, divided by (x + 7)(x - 8)³...this is all together.
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I have written this as a single logarithm, first by applying the power property,
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then by combining these two logs and these two logs, using the product property,
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and then finally combining this log and this log, using the quotient property.
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Now, we are going to apply what we have learned to actually solving logarithmic equations.
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And we have a technique for solving an equation, as long as we have a single log on each side.
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I need to combine this side, and I need to also get rid of this 1/2 out in front,
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so that I can have something of the form log < font size="-6" > b < /font > (x) = log < font size="-6" > b < /font > (y),
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and then use the property that y must equal x.
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First, I am going to apply the power property to get rid of these coefficients.
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log₉(2³) - log₉(2x - 1) = log₉(49^1/2).
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So, I can do some simplifying: I know that log₉(2³)...that 2 cubed is 8, so that is log₉(8).
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Now, 49^1/2 = √49, which equals 7; so I can write this as log₉(7)--it is already looking more manageable.
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On the left, I need to combine these; and I can do that using the quotient property.
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log₉...this is going to go in the numerator, and this will be the denominator.
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So, this is log₉(x) divided by 2x - 1 equals log₉(7).
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Once I am at this point, I have this situation, where I have the same base, and I only have one logarithmic expression on each side.
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So, I can just say, "OK, x equals y, so 8 divided by 2x - 1 equals 7."
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I am going to multiply both sides by 2x - 1, which is going to give me 8 = 14x - 7.
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I am going to add 7 to both sides to get 15 = 14x, and then I am going to divide both sides by 14: so 15/14 = x, or x = 1 and 1/14.
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It is important that I check the solution; so I want to make sure that I don't end up taking the log of a negative number.
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And the only log here that has variables as part of it is this one, so I am checking log₉(2x - 1), and I am letting x equal 15/14.
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log₉(2) times 15 times 14, minus 1...
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Now, instead of figuring out this whole thing and subtracting all of that, all I have to do is say,
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"All right, this is slightly more than 1; so 2 times slightly-more-than-one is going to be slightly more than 2."
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If I take a value of slightly more than 2 and subtract 1 from it, I am going to be fine; I will have a positive number.
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I will not be taking the log of a negative number, so this solution is valid.
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You don't have to get the exact value; you just have to check it far enough to be sure that what you have in here,
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what you are taking the log of, is not negative; it is greater than 0.
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OK, Example 4: we are asked to solve a logarithmic equation, and we almost have this form, but not quite.
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I can't use x = y, because I have this 1/2 here.
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In order to combine these two, they have to have the same base.
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In order for us to use the product property to combine these two logs into one, I have to somehow make 1/2 into log < font size="-6" > 16 < /font > .
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So, my goal is log < font size="-6" > 16 < /font > --that I am going to turn 1/2 into that.
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Well, let's think about this: if I think about my definition of logarithms, I have log < font size="-6" > b < /font > (x) = y.
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And what I want is log < font size="-6" > 16 < /font > of some number (but I don't know what number) is going to equal 1/2,
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because then, if I have this, instead of writing 1/2 here, I will just write this.
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Thinking of my definition of logarithms and how I can use that to solve an equation like this,
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an equation with a log in one side: now I have a separate equation that I need to solve in order to solve this one,
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just to substitute there: well, recall that log < font size="-6" > b < /font > (x) = y if b^y = x.
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So, I am going to solve this by converting it into its exponential form,
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which is going to give me 16^1/2 = x.
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This is the same as √16, which is 4; therefore, log < font size="-6" > 16 < /font > (4) = 1/2.
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These are equivalent; since these are equivalent, I can write that up there--I can substitute.
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Now, my next step is to combine these two logs on the right, using the product property.
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log < font size="-6" > 16 < /font > (12x - 21) = log < font size="-6" > 16 < /font > [4(x² - 3)].
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Now, I have it in this form, and I can say, "OK, 12x - 21 = 4(x² - 3)."
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And then, I just solve, as usual: this is going to give me 12x - 21 = 4x² - 12.
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I have a quadratic equation; I need to first simplify; so let's subtract 12x from both sides.
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I am going to set the whole equation equal to 0; that is -21 = 4x² - 12x - 12.
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I am going to add 21 to both sides to get 0 = 4x² - 12x + 9.
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I am going to rewrite this in a more standard form with the variables on the left.
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Now, this is just a matter of solving the quadratic equation; and this is actually a perfect square: it is (2x - 3) (2x - 3) = 0.
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And you can check this out to see that 2x times 2x is 4x²; 2x times -3 is -6x, plus -6x, is -12x; -3 times -3 is 9.
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According to the product property, if 2x - 3 equals 0 (and these are the same, so I only have to look at one factor), then I will have a solution.
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So, 2x = 3; if I solve for x, x equals 3/2.
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x = 3/2; now, my last step--this is a potential solution--is that I need to go ahead and check it right here.
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Let's let x equal 3/2; and we are going to insert that into log < font size="-6" > 16 < /font > (12x - 21).
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So, x equals log < font size="-6" > 16 < /font > (12(3/2) - 21)...not x equals...log < font size="-6" > 16 < /font > ...this cancels;
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this becomes 1, and this is 6; this is 6 times 3, minus 21; so this is log < font size="-6" > 16 < /font > (18 - 21).
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And that gives me log < font size="-6" > 16 < /font > (-18); therefore, this solution is not valid.
00:27:34.000 --> 00:27:44.600
And since that is the only solution I have, there is no solution to this equation.
00:27:44.600 --> 00:27:48.000
The only solution I came up with was an extraneous solution.
00:27:48.000 --> 00:27:55.800
I solved this by first converting it to this form; and I did that by setting up an equation
00:27:55.800 --> 00:28:02.600
where I figured out what x had to be for me to write 1/2 in the form log < font size="-6" > 16 < /font > .
00:28:02.600 --> 00:28:10.600
And I used the technique of converting this to the exponential form 16^1/2 = x and solving for x.
00:28:10.600 --> 00:28:19.000
That told me that log < font size="-6" > 16 < /font > (4) = 1/2; so instead of writing 1/2 here, I wrote log < font size="-6" > 16 < /font > (4).
00:28:19.000 --> 00:28:25.300
Then I used the product property to combine the two logarithmic expressions on the right.
00:28:25.300 --> 00:28:34.600
Then I used this property, and I set 12x - 21 equal to this expression, and solved, using factoring, a quadratic equation.
00:28:34.600 --> 00:28:40.200
But then, I went and checked my solution and found that it was not a valid solution.
00:28:40.200 --> 00:28:43.000
Thanks for visiting Educator.com; that concludes today's lesson.