Hi guys, welcome back to my channel.
So today we're going to be talking about expanding and condensing logarithms and we're going to also cover the change of base formula.
So let's take a look at our three basic rules.
First off we have the product rule.
So for the product rule
we have in the condensed form, that's where the name is coming from,
we have in the condensed form a product of m times n and when you expand it
it becomes a sum so still the same base log base b of m plus log base b of n.
Next, there's the quotient rule.
So in the condensed form
you'll see that there is a quotient m over n and then they expanded instead of having a sum
there is a difference. So log base b of m minus log base b of n.
And then the last rule is the power rule. So this is basically a repeated product rule if you have m times itself n times
you can view it as basically doing the product rule and times so you get a total of n
log base b of m terms. So let's again look at the name compared to where it comes from. So the product rule
comes from the product in the condensed form, the quotient comes from the quotient in the condensed form, and the power comes from the power.
One thing to note about the power rule is that everything inside
the argument has to be raised to the power.
Okay, so we are gonna have to be careful if some things are raised to power some things are not it's not just immediate
as some other things.
So let's go into our last rule, which is the change of base formula.
So the change of base formula isn't really helpful for condensing or expanding per se but it's the
rule that's gonna allow us to change any logarithm in a base that would be less desirable,
say 2 or 3, the bases that are less desirable tend to be bases that aren't particularly special.
So we tend to prefer bases like e and 10 over just other arbitrary bases.
So frequently, you'll be converting to base e and
how that goes is you start off with the base you had and
you think about the base you want.
So the formula that does it is you do log whatever base you are wanting to change to and
then of
the original argument divided by the log of that same base that you're trying to change to of
the old base.
Okay, so those are our rules that are at play and you'll notice there aren't any other rules, I'm not leaving any out.
So there's no rule that's gonna help us with if we have a sum inside the condensed form. We only have product, quotient, and power.
Okay, so let's take a look at for some practice problems here.
So expanding completely. How I like to do this is kind of a very global approach
I can kind of just look at this and break down some properties.
So when I'm expanding the first thing that I'll do is I'll isolate what is my numerator?
What is my denominator?
So the terms in the denominator are gonna go with the negatives and the
terms in the numerator are going to go with the positive log terms. So let's separate. Let's separately circle every single term.
Okay, so in the numerator, the term that I have is there's just one term root z and the denominator
I have two terms x and y so, I know that there's going to be three total log terms
and it looks like two of them are gonna be negative.
What goes in the negative ones is x and y and what goes in the positive one is
square root of z
for a first step, but notice I see a power. This is like having ln of z to the one-half and
everything inside that argument is raised to the power of 1/2 so I can use my power rule to bring that outside and that's 1/2
ln z minus ln x minus ln y and that's truly as far as we can go.
So when someone says expand completely they mean using those logarithm rules break it apart as much as possible.
This is going to be particularly helpful for
classes such as calc 2
because you'll need to be integrating things and it turns out it's actually much easier to integrate these logarithms when they're fully expanded.
Ok, so let's take a look at our next example.
Circling all our terms in the numerator, I have a 9, I have x minus 5 to the fifth, I have y plus 3.
Notice, these are terms in the product
not terms in a sum or difference. In the bottom I have y squared,
y minus 2, y plus 4.
Okay, so it looks like I'm gonna have a total of 6 log terms and the first three will be positive
so log base 3 of 9
plus log base 3 of x minus 5
to the fifth
plus log base 3 of
y plus 3 and then
on the bottom we have minus
log base 3 y squared minus
log base 3
y minus 2 minus log base 3
y plus 4.
Okay, and notice this is not a fraction, this is one long sum.
It's just I kind of ran out of room
so let me write in the next line, I will write it as one line where there's more space. And let's simplify as we go along.
So the first term log base 3 of 9 we can simplify that.
That's asking what power do you have to raise base 3 to to get 9 so that'll be 2 and
we have plus and it looks like the argument has been raised to the power 5 so we can use the power rule
to bring that 5 out
and
then we have plus log base 3 y plus 3 and then giving myself a little bit more room here
minus 2 times log base 3 of y
minus log base 3 of y minus 2 minus log base 3 of y plus 4.
And checking that all the powers have been handled,
yes, this power was handled, the 5 was handled, that 2 was handled.
Perfect! And
looking at what we have left this is truly as simplified as we can get.
So there's nothing that's going to enable us to get rid of the sums and differences inside logs.
Those have to stay there is just no log rule that
simplifies those down any further.
Ok, so moving on, we're going to go in the next direction.
Let's go instead of expanding, let's condense these logarithms completely.
So the first step I pretty much do the reverse of the process of expanding.
So the first step I'll do is I'll bring the powers inside and
I just bring the positive part of the power. Everyone
does this a little bit differently so it's okay if your way is a little bit different of seeing it as
long as we're getting to the same answer the way you're doing it, it's fine.
Okay, so we have ln x squared minus ln y cubed plus ln z to the fourth.
Now we're gonna try and put this into one logarithm and I know I'm gonna have a ratio because I see both
positive terms and negative terms. So let's underline what is gonna go in the numerator and what's gonna go in the denominator.
So it looks like positive the ln x squared term is positive and the ln of Z to the fourth is positive
versus the
ln y cubed is negative. So when I write my fraction, I'm gonna have in the numerator x squared,
z to the fourth and then in the denominator
goes the y cubed. And then I notice my fraction cannot be reduced any further. So that's
our answer.
Cool okay. So let's take a look at this next one. We got a little challenge mode here. So
check out the bases.
We've got two and we've got a half and if we scroll up and look at our log rules
notice that the pattern for condensing is you need to have
the same base.
So I cannot immediately apply any of these rules: product, quotient, or power really at this point.
I have to try and first change the base.
Okay, now you might be wondering why would we want to do this at this point.
Well, just like with integrating you want to have expanded logarithms, when you're solving log equations
you typically want to have condensed logarithms in the same base.
So I wouldn't want to solve an equation that involved this sum without first changing the base to two.
So I'm going to change this log base 1/2 of x minus 1.
Remember the pattern is you do log base of what you want. So I want 2,
log base 2 on top and bottom and
then
you do log base 2 of what was the original argument over what was the
old base.
So log base 2 of 1/2 and now we need to simplify that log base 2 of 1/2
that's asking what power do you raise the two to get a 1/2? And the answer would be negative one.
So it looks like this simplifies as
log base two of x
minus log base two of x minus one and now I see that I have
one positive term,
one negative term so when I write the formula for my condense logarithm
I'm gonna have x in the numerator and x minus 1 in the denominator.
And I can't reduce that fraction any further
so
there she is.
Okay one note before we finish this video for today and that note is when you're unsure about how to simplify something
it is better to not simplify it.
Especially in an exam setting so at home
you can obviously just look up whether what you're simplifying is valid or you can take your time and check it carefully
but when you're in a timed situation such as an exam
this is very much my philosophy some teachers may tell you something different. But personally, I'd rather see someone work with something
that's unsimplified but correct then try and simplify
using false rules of mathematics and end up with something
that's just not equivalent to what they're working with and get wrong answers from that point in the problem forward.
So frequently in calculus, you're gonna see problems that build and you'll need to take answers that you get and work with them further.
So that's where it gets a little dangerous
simplifying when you just don't know how so I'll just leave you guys with a note for me, please,
don't simplify unless you're a hundred percent confident that that's how it simplifies. Otherwise, leave it as is.
Hopefully, this video was helpful and tune in next time and we'll see how to solve some exponential and logarithmic equations.
