WEBVTT
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Hi, let's take a look at solving some logarithmic equations.
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The technique used to solve a specific log
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equation really depends on what the original
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problem looks like, so some analysis is
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always in order before you begin.
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Let's take a look at an example: log base 3 of 2 x minus 1 equals 2.
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That's what we want to do, we want to solve
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for x, so let's analyze what we have.
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Well, we have a log expression: we have a
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log, a base, we have this term, which is
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called the argument, and then
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we have this number out here, 2,
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which is actually the logarithm.
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Now think about the definition of logarithm.
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So, we have log base a of x equals y,
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and this means that a to the y equals x.
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Now there are some restrictions on these variables.
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First of all, a cannot be 1. What else?
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a has to be greater than 0, and then there
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is one other one, and this is probably the
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most missed part of a problem like this.
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What sits in this position, whether it be
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just x like it is here in the definition or
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2 x minus 1 as it is in our sample one, has to be greater than 0.
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So, the first thing to do is to look at this
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problem and realize the 2 x minus 1 has to
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be greater than 0, and so this is the domain of this expression.
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Alright next, let's take a look at the form of what we have.
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We have a log, a base, an argument, and a
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number, and this exactly matches log, base,
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argument, number, which means all we have to
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do in order to rewrite this is to follow the definition, and then
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of course we have to remember at
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the end that we're going to have to check to
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be certain that the value we get for x,
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when put back into here, makes this expression greater than 0.
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Okay, so we rewrite 3, which is the base a,
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to the second power equals 2 x minus 1, and
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now we have a linear equation, and we solve for x.
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So, we have 9 equals 2 x minus 1. 2 x equals 10, and so x is 5.
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I switched the sides of the equation.
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You don't have to do that, but it's just
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traditional to have the variable on the left.
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So, we got x is 5, are we done? No, we have to check.
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We want to check to be certain that this
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value, when placed into here, makes this
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expression positive. So, what would we have?
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We have 2 times 5 minus 1, which is 9, which
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is definitely greater than 0, and so this is
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a valid solution to this first log example.
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Okay, now let's take a look at another example.
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In this example we have the expression 2 log
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base 4 of x equals log base 4 of 25.
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First thing I notice when I analyze this is
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that there's a coefficient, 2, in front of
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this expression, and so I recall a property
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that's going to allow me to rewrite this.
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Okay, can you think of what the property is?
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Well, here it is, it looks like this: if you
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have log base a of some expression, x,
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raised to a real number power, you can
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rewrite this as r log base a of x.
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Okay, in this case, we have the right-hand
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side, we're going to rewrite it as the left-hand side.
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The same restrictions hold here: a cannot be
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1, a has to be greater than 0, and again
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what sits in this position has to be positive.
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Okay, using that property, the first thing
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I'm going to do is rewrite this as log base 4 of x squared.
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The 2 is written here, and then I just copy over the right-hand side.
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Alright, now what do we have? Go back and look at example one.
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Does this look like the same format as example one?
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Not really, we have a log expression on the
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left, but in example one, we had a constant
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here, and in this example, we have another log expression.
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So, instead of using the definition to
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rewrite this and solve, we have to use the
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one to one property of logarithms. So, let's go back over that.
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Well it looks like this: if you have the log
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base a of some expression, and we'll just
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call it capital M and in this case that's
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x squared, and it equals the log base a of
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capital N, we're going to put an if in front of this.
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If this is true, then what do you think is true?
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Well, you're right: M equals N.
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Now of course, there are restrictions here
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again: M, N, and a all have to be positive and again a cannot be 1.
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Alright, so how do we use this property?
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Well we look at this and we see that the
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bases are the same: a, a, in this case 4 and 4, and so what do we do?
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We realize that the one to one property,
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which is what this is called, allows us to
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very simply reduce this expression to x squared equals 25.
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Okay let's see, we know that we can use the
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square root property now and get plus or minus 5,
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and it's tempting to stop here, but don't.
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Always remember that when you use or work
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with a log expression at all, anything you
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do, this has to be greater than 0, and so does this.
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Okay, well the 25 is already greater than 0,
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and so now all we have to do is check this. So, what happens?
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Well, when we check positive 5, of course
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positive 5 put into here is greater than 0, and so that's fine.
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But negative 5 put into here is not greater
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than 0, and so that root does not check.
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So, what we have here is actually one root:
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positive 5, and that is the solution to this example.
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Okay now let's look at what happens when the
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log expressions get a little more complicated.
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In this third example, we have the sum of
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two log expressions and then a constant, so
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this example looks a little different in
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terms of the format of the first two that we did.
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So, let's analyze this. What could we possibly do to simplify this
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to get it into a form where we could easily
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use either the process we did for example
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one, where we use the definition, or the
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process we did for example two, where we use
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the one to one property? Well, you're right.
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What we can do is we can add these two
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expressions, so let's review how this is done.
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If you have the log base a of an expression
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added to the log base a of another
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expression, then yes, you can rewrite this
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as the log base a of M times N.
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Okay, same restrictions: a, M, N have to be
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greater than 0, and what's the other one, a cannot equal 1.
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So, let's check: a is 5, this is M, this is N, and so what can we do?
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We can combine this expression, so we get
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log base 5 of M times N, so that would be
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x plus 6 times x plus 2 equals 1.
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Alright, did we improve the situation?
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Yes, we did because we have to multiply
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these two terms in order to eventually
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solve, but what's important about what we
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did is we got this in standard form.
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So, we have log, base, argument, and then
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the constant here, which means we can go
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back to the definition now, like we did in
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example one, where we have log base a of x
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equals y means a to the y equals x.
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Same restrictions: x of course has to be greater than 0.
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Okay, so let's do that.
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As I copy this over, I'm going to multiply M
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times N so I'm going to get x squared plus
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8 x plus 12, that's this expression,
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and then I'm going to rewrite the logarithm
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in exponential form, so I would then have 5 to the 1.
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Now in this case, I have these two sides
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reversed, but this is an equality, and
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that's fine, and we usually find it more
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convenient to have the x on the left.
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Alright, so let's solve this quadratic
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equation: x squared plus 8 x, subtract 5
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from both sides, and then let's see
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if this factors easily. I believe it does.
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Okay, we set each factor equal to 0, and we
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get x equals negative 1 and x equals negative 7.
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Are we finished? No, we have to check.
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We have to check both roots, and not only
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both roots but both roots in this expression
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and in this expression. If we find one
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situation where negative 1 does not produce
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a positive argument here, then that is not a root.
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Alright, so let's check the negative 1 first.
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Okay, when I put negative 1 into here,
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negative 1 substituted here would be 5, so
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that's greater than 0. That's good.
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Negative 1 here would be negative 1 plus 2, which is 1.
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That means this is a valid root.
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Even though it's a negative number,
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it's not x that has to be positive, it's the entire expression.
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Okay, let's check negative 7.
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Did we just assume that it works or not works?
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No, you always have to check it.
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Alright, so negative 7 into here, what happens?
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Right away, you see that you have negative 7
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plus 6 which is negative 1, and that is
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clearly not greater than 0. So this root does not check.
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Let's go back and recap what we did here.
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First, we used this law property to combine the two log terms.
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Then, when we had this in standard form, we
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used the definition of logarithm, just like
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we did in example one, to rewrite this in
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exponential form, solve for x, and then we always check.
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Okay, let's look at one more example.
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In our fourth example, let's look at what we have.
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First of all, try to remember what does this mean?
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This is log base e or natural log, so we
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have four natural log expressions.
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What can we do to solve for x?
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Well, let's see if we can condense some of this.
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Remember the property that looks like this:
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if you have log base a of M minus
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log base a of N. How can you combine these two expressions?
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Rewrite it as this, exactly, same
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restrictions as always in a, M, and N.
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Okay, we're going to be able to use this
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property on the left side and the right side.
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Okay, we're going to rewrite this.
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Bases are the same, so I'm going to rewrite
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this as natural log of x which is M divided
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by x plus 1, and on the right-hand side the
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natural log of the quotient of x plus 3 and x plus 5.
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Alright, we've simplified it, what do we do now?
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Well, we generally have two things we can
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use: we can use the definition, which is not
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really going to be that helpful here, or we
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can go back to that example two and the one to one property.
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So, we have log base e of an expression
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equals log base e of an expression.
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Okay, so let's go back and look at that
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property: log base a of M equals log base a
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of N, and if this is true then M equals N,
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which means that what I can do is I can
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rewrite these two expressions in this form.
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Do a little algebra, solve for x, and remember to check.
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Okay, let's cross multiply.
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So, x times x plus 5 would be x squared plus 5 x.
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Multiplying across this way, I would have x squared plus 4 x plus 3.
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Okay, well look what happens: x squared and
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x squared, and so I have 5 x equals 4 x plus 3,
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so I subtract 4 x from both sides, and I get x is 3.
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Alright, we still have to check.
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Once again, what are we checking for?
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We are checking to be certain that this value is in the domain of
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all four of these expressions: this one, this one, this one, and this one.
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Well, let's see what happens. Three is positive, so 3 here checks
000:16:34.500 --> 000:16:37.070
positive, greater than 0; 3 plus 1 is 4,
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that's also greater than 0 so that checks;
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it's going to check here, 6, and it's also going to check here, 8.
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So, what does that mean? We have the solution, and we now have a
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solution that has been checked.
000:16:56.171 --> 000:17:00.421
These four examples are examples that you should go back and study.
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You should rework them on your own, then you should go back and do
000:17:04.672 --> 000:17:06.921
similar exercises for practice.
000:17:06.921 --> 000:17:13.170
That's what it will take to build skill solving logarithmic equations.
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Let's take a look at solving some exponential equations.
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The technique that is used to solve a
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specific exponential equation really depends
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on what the original problem looks like, so
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some analysis is always in order before you begin.
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For example, in this exponential equation,
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we have 3 to the x plus 1 equals 1 over 81.
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The x is in the exponential position, and so
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what we have to do is figure out how we should approach this problem?
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The simplest thing to do here is to realize
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that 81 can be rewritten as 9 times 9 or
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3 to the 4th, which means 1 over 81 can be
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rewritten as 1 over 3 to the 4th, which can
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then further be rewritten as 3 to the negative 4.
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This means that with very little effort
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I can rewrite the original problem so that I have like bases.
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When this happens, I can use a one-to-one property. Let's refresh.
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If I have a to the M equals a to the N, then what is also true?
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Right, M equals N. Here, as always,
000:18:54.391 --> 000:19:00.642
the base a cannot be 1 and has to be greater than 0.
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So, what can we do now?
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Well, we can set the two exponents, the M
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and the N, equal. Solve for x, and we get x is negative 5.
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We're not required to check this as we are
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when we're solving logarithmic equations to
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check for whether or not the value is in the
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domain, but it's always a good idea. Let's just check.
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Negative 5 into here would be negative 5
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plus 1, so this would be 3 to the negative 4,
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which we know is 3 to the negative 4.
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Working with a problem of this type is
000:19:44.541 --> 000:19:47.642
called solving an exponential equation with like bases.
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This only applies for certain problems that can be rewritten like this.
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So, let's take a look at what happens when
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you can't rewrite an expression with like bases.
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In this example, 2 to the x equals 5.
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We will not be able to rewrite each side in
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terms of the same base, so we need to look for another method.
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Recall the definition of logarithms that
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allows you to change from exponential form
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to log form: a to the y equals x means log base a of x equals y.
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Be careful when you do this: these are
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positional notations, and so this x in this
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position is not the same as this x here.
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So, what we actually have is: this is the 2,
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this is the x, and this is the 5. So, let's rewrite in log form.
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So, it would be log base 2 of 5 equals x.
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Now this is an exact answer, but if you
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needed to evaluate this on a scientific
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calculator, you would have to have it in
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either base e or base 10, so let's convert this.
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Right now, it's in base 2. We have the change of base formula: a log
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base a of b can be rewritten as the natural
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log of b divided by the natural log of a.
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That's using base e, or you can use base 10: log of b divided by log of a.
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So, let's use natural log for this.
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I would rewrite this expression as the
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natural log of 5 divided by the natural log
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of 2, and this is an exact value for x that
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could be easily evaluated on a scientific calculator.
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Let's take a look at another example.
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In this example, 4 times 3 to the x plus 1 equals 6.
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We want to solve for x. The first thing to notice is that this
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exponential equation is not in what we consider standard form.
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This 4 is not part of the base, the base is 3,
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so the first step is going to be to rewrite this in standard form.
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I'm going to divide both sides by 4, and so
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I get 3 to the x plus 1 equals 6 over 4.
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Now that it is in standard form, what method should we use?
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Because we have a base, an exponent, and
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a number, we can simply rewrite this as a log expression.
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Remember that, if you have a to the y equals
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x, that this means log base a of x equals y.
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So, from this example, the a is the 3, the y
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is the x plus 1 term, not to be confused
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with this x term, and then the 6/4 is x.
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So, when we rewrite in log form, we would
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have a 3 here, we would have 6/4 here,
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and x plus 1 out here. So, let's do that.
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Log base 3 of 6/4 equals x plus 1.
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We want to solve for x, so the next thing to
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do is to subtract 1 from both sides.
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We would have log base 3 of 6/4 minus 1, and
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you can see that we're going to need
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parentheses here because we want it to be
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perfectly clear that the 6/4 is part of the
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log, but the 1 is not, and we get x.
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This is an exact value, but it is not in
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either base e or base 10, which is where we
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would want this if we had to evaluate.
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So, we look at the change of base formula:
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log base a of b can be rewritten as the
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natural log of b divided by the natural log
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of a or log base 10, so it would be log base 10 of b divided by log base 10 of a.
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I'm going to use natural log, and so I could
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rewrite this expression in parentheses as
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the natural log of 6/4 divided by the
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natural log of 3 and then minus 1 equals x. This is the exact value.
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If you needed to evaluate this, you could do
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this on a scientific calculator.
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Now let's take a look at a more complicated example.
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In this fourth example, we have 5 to the
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x minus 2 equals 4 to the 3 x plus 1.
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Right away, you can see that because you
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have variables on both sides in the
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exponential position, you're not going to be
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able to use the definition immediately to simplify this.
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So, instead what we're going to do is use
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the following property: if M equals N, now
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let's look at what an M is, M is this entire
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expression, N is this expression, the two
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expressions are equal, then we can rewrite
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this as a log base a of M equals the log base a of N.
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I would recommend either using natural log
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or log base 10 because you may, at some
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point, want to get an approximate answer,
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and that's what your scientific calculator has.
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So, applying this property and using natural
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log, on the left side I would get the
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natural log of 5 to the x minus 2, and on
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the right side the natural log of 4 to the 3 x plus 1.
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Okay, what have I done? Have I made this any easier to solve?
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Well, it looks longer, but as it turns out,
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we can do something else at this point that
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will make it much simpler to solve.
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We can recognize here that if we have the
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log base a of x to the r, so that's log base
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e of 5 to the x minus 2, where the x minus 2
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is the r, we can rewrite this as r times log base a of x.
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That's one of the properties.
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So, what we're going to do is take this
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expression and rewrite it with the r in the
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front, in this case x minus 2 is the r.
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So, we have x minus 2 times the natural log
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of 5 and 3 x plus 1 times the natural log of 4.
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The natural log of 5 is a constant.
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Let's distribute it, so we get x ln 5, and then minus 2 ln 5.
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On the right-hand side, we distribute ln 4
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here and here, so we get 3 x ln 4 plus 1 ln 4.
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Gathering the x terms on the left,
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I subtract 3 x ln 4 from both sides.
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Copy this expression, and add 2 ln 5.
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Alright, let's look for a minute.
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So, 3 x ln 4, because it has an x, has been subtracted from both sides.
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This term does not have an x: it has been
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added to both sides and is now here.
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We want to factor out the x because that is what we are trying to solve.
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So, when we factor out the x, what remains is ln 5 minus 3 ln 4.
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One times ln 4 can be rewritten as ln 4, and then here we have 2 ln 5.
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Dividing both sides by what is in
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parentheses, I get x is the natural log of 4
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plus 2 times the natural log of 5,
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all divided by the natural log of 5 minus 3 natural log of 4.
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This is the exact value of x using all of the
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original numbers from the original problem.
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There are things you could do to simplify
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this, including using your calculator to get an approximate answer.
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The best thing to do at this point is to
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read the directions and decide what you are
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trying to do in terms of the form of your answer.
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You can see from these four examples that
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there are varying techniques used to solve exponential equations.
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Study these examples, practice them on your
000:31:34.122 --> 000:31:37.124
own, and then practice similar examples
000:31:37.124 --> 000:31:43.500
until you have achieved a high level of skill solving these equations.