WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today we are going to talk about base e and natural logarithms.
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And these are frequently used in real-world applications.
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So, first, the number e: the number e is an irrational number that is approximately equal to 2.71828; it is called the **natural base**.
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Exponential functions that use the base e are called natural exponential functions.
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And these are used to model processes that grow or decay.
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For example, something like population growth might involve the use of base e.
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You can treat these like other exponents and other exponential equations, just recalling that it has a specific value.
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And so, it is structured similarly to other exponential functions.
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For example, f(x) = e^3x - 1 could be an exponential function using base e, or something like f(x) = 4(e^x).
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When we work with logarithms to the base e, these are written with a special notation, ln(x).
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We don't use log(x) in this case; we use ln(x); and this is called the **natural logarithm**.
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So, when you see ln(x), really, the base is e.
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But we don't write it that way; it is just written as ln(x).
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And this is the inverse of the function f(x) = e^x.
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Therefore, since these are inverses, you end up with the identity functions.
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Again, this is a composition of functions; if I am going to let...let's let f(x) equal ln(x), and I am going to rename this g(x) to keep things clear.
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So, according to composition of functions, if these two are inverses, f composed with g should give me back x.
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So, f composed with g would be f(g(x)), which would then equal f(e^x); and this is ln(x), so it would be ln(e^x).
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And the natural logarithm of x, and e to the x power, essentially cancel each other out, and we just end up with x, the natural log and this base e.
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g composed with x should do the same thing--should give me g(f(x)), which is going to equal g(ln(x)), which equals e^ln(x).
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And the base e and the natural log will cancel out; I will end up getting x back (it is the identity function).
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And we are going to use this property as we solve equations involving base e and natural logs.
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Briefly reviewing what the graph is going to look like: if I look at the graph of f(x),
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this is going to look like the typical graph of a logarithmic function,
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where it is going to increase as x becomes more positive, and the y-axis is going to function as an asymptote.
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Therefore, this graph is going to approach x = 0, but it is never actually going to reach it; and so that domain is going to be restricted to positive numbers.
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Whereas, if we are looking at the graph of g(x), this is going to look like the graph of a typical exponential function.
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And it is going to increase: as x becomes positive, y will increase, and here the x-axis is functioning as an asymptote.
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So, g(x) = e^x.
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When we are working with equations and inequalities involving powers of e, we can proceed as follows.
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If we are working with the powers of e in an equation or inequality, we are going to use natural logs to solve them.
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We talked earlier about using logarithms to solve exponential equations that have different bases.
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We would find the common log of both sides; here, if the base is e, then you use the natural log.
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If you have an equation or inequality that involves natural log, you can use powers of e.
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If you have powers of e, use natural logs to solve; if you have natural logs, use powers of e.
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All the properties we discussed previously for logarithms are valid (like the product property, the quotient property, and the power property).
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In addition, when you are working with logarithms, you always have to be careful for extraneous solutions.
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So again, we are going to talk about extraneous solutions and make sure that we are looking out for those
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and that we check for those before we decide that the solution that we found is valid.
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For example, if I am working with 2^x + 6, and that equals 7^3x - 2,
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earlier on in the course, we said that we would just find the common log of both sides;
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and that is going to give me log₇(3x - 2).
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And then, from there, we went about isolating x and went on down.
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We are doing the same thing here; only now we are going to talk about natural logs.
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For example, e^5x - 3 = 9: since I am talking about powers, I am going to use natural logs to solve.
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So, I am going to take ln(e^5x - 3) = ln(9), the natural log of 9.
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Remember that using the inverse property, if I am taking ln(e to some power), I am going to get x back; this is the identity function.
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That is very helpful, because I will just end up with 5x - 3 = ln(9).
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This is a number with a specific value; again, you can use your calculator to find it.
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It is so commonly used that there is going to be a button on your calculator for this.
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Now, I just need to isolate the x; 5x = ln(9) + 3; x = ln(9) + 3, divided by 5: x will be isolated.
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We started out with powers of e; and we used natural logs to solve.
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Conversely, if I am given natural logs, such as ln(x + 3) = 4, I can take both sides and make them base e, and raise them to the power...
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here it would be ln(x + 3); on the other side, I am going to get e⁴.
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Again, I have inverses: e^ln or ln(e) were inverses; these canceled out; I got 5x - 3.
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The same here: e^ln(x + 3)...I end up with just x + 3 = e⁴.
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Therefore, x = e⁴ - 3; and this is a solution, because we know that e is a specific value (a little more than 2).
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We take that to the fourth power and subtract 3 from it, and we have a specific value for x.
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I did start out with a logarithm, though; so I have to make sure that I am not ending up with a solution that will give me something negative.
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Let's go ahead and look back to check that: ln(x + 3)...and I am letting x equal e⁴ - 3.
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So, ln(e⁴ - 3 + 3): this is ln(e⁴, because these cancel out.
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And I know that e is positive, that it is greater than 2; so I take that to the fourth power; that is positive as well.
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So, this is a valid solution.
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In the first example, I have 4 times e^4x = 40; so I am working with a power of e, so I am going to use natural logs to solve.
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To make this simpler, I am going to first divide both sides by 4.
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Next, I am going to take the natural log of both sides.
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And, according to my inverse property, ln(e^x) = x; because these are inverses,
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the identity function is going to give me ln(e^x) simply equals x.
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Therefore, I end up with 4x = ln(10), the natural log of 10.
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I am going to isolate x now by dividing both sides by 4; so x = ln(10), divided by 4.
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Again, as soon as I saw that I had a power of e, I knew that I could solve this using natural logs.
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First, I am just dividing by 4 to make this simpler, taking the natural log of both sides, and then realizing that,
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because these are inverses, I am going to end up with 4x = ln(10), and isolating the x.
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This is a logarithmic equation where I can use the properties of logarithms to solve,
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just as I did with earlier equations when we used other logs, base 10 logs, or other bases.
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Remember that I can solve these as long as I get them in some kind of form.
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Earlier we talked about log < font size="-6" > b < /font > (x) = log < font size="-6" > b < /font > (y) if x = y.
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Well, if I get ln(x) = ln(y), these have the same base, so x = y as well.
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Even though we use a different notation, ln, it is still just a log, only the base happens to be
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a certain number that we are specifying, which is e, worth a little bit more than 2.
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All right, so we are going to start out using our regular properties for logarithms to get this into a form that we can work with.
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First, I have ln(x² + 12) - ln(x) - ln(8).
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I want to combine these first two; and I can do that using the quotient property.
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because ln(x² + 12), divided by x...all of this can be combined,
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because if I am subtracting one log from another (I have the difference of two logs), I can just take the log of the quotient of those two.
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Again, I have subtraction; so I want to use the quotient property once again.
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This is going to give me ln(x² + 12, divided by x, divided by 8).
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So, I use the quotient property again, because I had all this divided by this.
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I want to simplify this complex fraction; so if I am dividing this numerator by 8, that is the same as multiplying this by 1/8,
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which will give me ln(x² + 12, divided by 8x).
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I just took this and multiplied by 1/8; that is going to give me x² + 12, divided by 8 times x, or 8x, equals 0.
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Now, I got this far; and looking at this, I actually only have a log on one side.
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So, this isn't going to work; so what is my other option?
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Recall that, when I am working with natural logs, the easiest thing to do is going to be to take a power of e.
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So, our strategy, if we have logs on two sides, is to use this.
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If we have a log only on one side, the strategy is going to be to take the equivalent exponential expression.
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So, I need to take both sides and make them e and raise them to this power and this power.
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Let's go up here to the second column and continue on.
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This is going to give me e^ln(x² + 12, divided by 8x) equals e⁰.
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I have inverses here; so they cancel out, and that leaves me with (x² + 12)/8x.
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Any number to the 0 power is simply going to be 1; it doesn't matter if it is 4 to the 0 power, or e (which it happens to be--a little bit more than 2).
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It is going to end up being 1: now I have something that I can solve.
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x² + 12 equals...multiply both sides by 8x; I see I have a quadratic equation.
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This is x² - 8x + 12 = 0; now I can factor; I have a positive here and a negative there,
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because this is...actually, I have two negatives, because this is a positive, and this is a negative; so I have two negatives.
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And my factors of 12 are going to be 1 and 12, 2 and 6, and 3 and 4.
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And I know that 2 and 6 together...2 + 6 equals 8, so I am going to make this 6 and this 2.
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That is x²; that is -2x + -6x gives me -8x; -6 times -2 gives me positive 12; so I factored correctly.
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Using the product property, I know that x - 6 = 0; that would be a solution, so that is going to give me x = 6.
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Or, if x - 2 = 0, I have a solution; and that gives me x = 2.
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Now, I have two solutions; but I always want to make sure that they are valid solutions, since I started out working with logs.
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So, I am going to go back up here and check them: ln(x² + 12) = ln...I will check the first one:
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that is going to be 6 squared plus 12, and that is going to be 36 + 12ln(48); that is valid.
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The second...I need to check 6 over here; ln(6); that is also going to be valid, because that is taking the log of a positive.
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Now, let's check 2: ln(x² + 12) = ln(2² + 12) = 12 + 4, so that is ln(16).
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That is a positive number that I am taking the log of, so that is valid.
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I am also going to check it here: ln(2) is positive; so both of these are valid solutions.
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Both of the solutions I have are valid: x = 6 and x = 2.
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Now, I am working with the power of e, so I am going to use natural logs to solve this.
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But first, I am going to make it a little bit simpler by adding 9 to both sides.
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So, I have an inequality; but I don't need this extraneous 9 on the left making things difficult, so e^-3x < 12 + 9, which is 21.
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I am going to take the natural log of both sides: ln(e^-3x) < ln(21).
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These are inverses; they essentially cancel each other out, leaving me with -3x < ln(21).
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I am going to divide both sides by -3, and I am going to immediately switch the inequality symbol to > ,
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because I am dividing by a negative number, dividing both sides by -3.
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So, it is going to give me -ln(21), divided by 3; so x > -ln(21)/3.
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Again, I first just added 9 to both sides and made this a little bit simpler.
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And then, I recognized that I am working with powers of e, so I can solve this by taking the natural log of both sides.
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OK, here I have an inequality that involves logarithms; and it is the natural log, so I am going to keep in mind that I can use powers of e to solve this inequality.
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First, let's just go ahead and divide both sides by 3 to make this a bit simpler, combining the constants on the right.
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Now, I have ln(2x - 3) ≥ 8: I am going to take powers of e of both sides.
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That is ln(2x - 3) ≥ e⁸.
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Since these are inverses, they essentially cancel each other out--they negate each other--so I get 2x - 3 ≥ e⁸.
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Now, all I have to do is solve for x: 2x ≥ e⁸, and add 3 to both sides, so this is going to give me e⁸ + 3.
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Then divide both sides by 2: x is going to be greater than or equal to (e⁸ + 3)/2.
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But I am working with logs, so I need to check back and make sure I don't end up with something that is going to make this negative.
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So, this says that x is greater than or equal to this; so the smallest this value of x will be is this.
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So, if I check it for this, and it is OK, all of the larger values will be OK, as well.
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ln(2x - 3): let's let x equal e⁸ + 3, divided by 2; and if this is OK, the whole solution set is OK.
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This gives me ln(2((e⁸ + 3)/2) - 3).
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This looks worse than it actually is, because the 2's cancel out: this gives me ln(e⁸ + 3 - 3).
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These cancel out; this gives me ln(e⁸).
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And I know that, by the identity property, this is actually going to be 8, so I am fine.
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I could also just say that I know that e is a specific number, and it is 2.
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And so, if I take a number slightly more than 2 to the eighth power, I am definitely going to get something positive.
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So, I am going to be taking the natural log of a positive number; and then these other values are going to be even greater, because this is the minimum value.
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Today, we talked about natural logs and the base e, and talked about solving equations that involve powers of e, as well as natural logarithms.
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Thanks for visiting Educator.com; I will see you next lesson!