Main content

## Algebra 2 (Eureka Math/EngageNY)

### Unit 3: Lesson 5

Topic B: Lessons 10-12: Logarithm properties- Intro to logarithm properties (1 of 2)
- Intro to logarithm properties (2 of 2)
- Intro to logarithm properties
- Using the logarithmic product rule
- Using the logarithmic power rule
- Use the properties of logarithms
- Using the properties of logarithms: multiple steps
- Proof of the logarithm product rule
- Proof of the logarithm quotient and power rules
- Justifying the logarithm properties

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Intro to logarithm properties (1 of 2)

CCSS.Math:

Sal introduces the logarithm identities for addition and subtraction of logarithms. Created by Sal Khan.

## Want to join the conversation?

- What is the real life use for Logarirhims? Why would people need them?(125 votes)
- I would add computer science to this list. The notations of the algorithms often use logarithms:)(7 votes)

- Why is this piece of math method called logarithm?(41 votes)
- Actually, it doesn't seem to be a contraction of "logical arithmetic". The Online Etymology Dictionary says (http://etymonline.com/index.php?term=logarithm&allowed_in_frame=0) "literally "ratio-number," from Greek logos "proportion, ratio, word" (see logos) + arithmos "number" (see arithmetic)". Wikipedia says (https://en.wikipedia.org/wiki/Logarithm) "Napier [...] introduced the word "logarithm" to mean a number that indicates a ratio: λόγος (logos) meaning proportion, and ἀριθμός (arithmos) meaning number".(26 votes)

- What happens if another number is added to the equation?

If you don't know what I mean, here is an example:

log2 A + log2 B + 3

Note that 2 is the base.(17 votes)- That expression could be simplified to

log₂(AB) + 3

But note that 3 = log₂(2³), so we can substitute that:

log₂(AB) + 3 = log₂(AB) + log₂(2³)

= log₂(AB) + log₂(8)

= log₂(8AB)

So, the general rule would be:

logₐ (b) + c

= logₐ ( baᶜ )(50 votes)

- Is it possible to have negative logarithms when dealing with properties. Let's just say log^2 -35 or something similar where you can expand it?(11 votes)
- Yes, the logarithms that have negative numbers as their argument are called complex logarithms, However it would be hard to handle them because their values can vary.

http://en.wikipedia.org/wiki/Complex_logarithm(9 votes)

- So, logarithmic addition is a process of exponential multiplication?(3 votes)
- Yes, because like solving problems that don't represent exponential properties, things can be simplified.(3 votes)

- Why did you put 81 as 1/81 instead of 81/1 if we started with 81 as a whole number?(7 votes)
- He did that because multiplying 1/9 *1/81 is the same as 1/9 divided by 81. Multiplying by 1/81 is easier to work out than 1/9 divided by 81. Always remember: dividing by a number is the same as multiplying it by it's inverse.

Example:

10/2 is the same a 10*1/2=5

20/4 is the same as 20*1/4=5

If you want to multiply instead of divide, just take the inverse or reciprocal of the number you want to divide by.(10 votes)

- Why can we assumed log1 as log base10 of 1?(6 votes)
- Because when there is no base specified that's an automatic 10.

This is what we call a common logarithm

That's why......

log 1 = log (base 10) 1

Note: A common logarithm is a logarithm with a base 10 so that's....

log (base 10) R=P or simply written as log R=P

where:

R= is the resulting number after raising a base to a certain power.

P= power(9 votes)

- What is the difference between log and ln?(4 votes)
- Khan Academy observes the convention that "log" means logarithm to base 10 (log₁₀, AKA the common logarithm), while "ln" is the logarithm to base 𝑒, which is also called the "natural logarithm."(6 votes)

- So what if the problem is Log2 = 6?(0 votes)
- this isn't a problem, this is a wrong statement.(33 votes)

- Can someone explain how log_a b = 1/log_b a?(3 votes)
- Actually, log_a b = log_b a is usually not true. For example, log_10 100 = 2 but log_100 10 = 1/2.

The relationship is instead log_a b = 1/(log_b a).

Let x = log_b a. Then b^x = a, and so a^(1/x) = (b^x)^(1/x) = b^(x * 1/x) = b^1 = b.

So log_a b = 1/x = 1/(log_b a).

Have a blessed, wonderful day!(2 votes)

## Video transcript

Welcome to this presentation
on logarithm properties. Now this is going to be a
very hands-on presentation. If you don't believe that one
of these properties are true and you want them proved, I've
made three or four videos that actually prove
these properties. But what I'm going to do
is I'm going to show you the properties. And then show you how
they can be used. It's going to be
little more hands-on. So let's just do a little
bit of a review of just what a logarithm is. So if I say that a-- Oh
that's not the right. Let's see. I want to change--
There you go. Let's say I say that
a-- Let me start over. a to the b is equal to c. So if we-- a to the b
power is equal to c. So another way to write this
exact same relationship instead of writing the exponent, is
to write it as a logarithm. So we can say that the
logarithm base a of c is equal to b. So these are essentially
saying the same thing. They just have different
kind of results. In one, you know a and b and
you're kind of getting c. That's what exponentiation
does for you. And the second one, you know
a and you know that when you raise it to some
power you get c. And then you figure
out what b is. So they're the exact same
relationship, just stated in a different way. Now I will introduce you
to some interesting logarithm properties. And they actually just fall
out of this relationship and the regular exponent rules. So the first is that the
logarithm-- Let me do a more cheerful color. The logarithm, let's say, of
any base-- So let's just call the base-- Let's
say b for base. Logarithm base b of a plus
logarithm base b of c-- and this only works if we
have the same bases. So that's important
to remember. That equals the logarithm
of base b of a times c. Now what does this mean
and how can we use it? Or let's just even try it
out with some, well I don't know, examples. So this is saying that-- I'll
switch to another color. Let's make mauve my--
Mauve-- I don't know. I never know how to
say that properly. Let's make that my
example color. So let's say logarithm of base
2 of-- I don't know --of 8 plus logarithm base 2 of-- I
don't know let's say --32. So, in theory, this should
equal, if we believe this property, this should equal
logarithm base 2 of what? Well we say 8 times 32. So 8 times 32 is
240 plus 16, 256. Well let's see if that's true. Just trying out this number and
this is really isn't a proof. But it'll give you a little bit
of an intuition, I think, for what's going on around you. So log-- So this is-- We
just used our property. This little property that
I presented to you. And let's just see
if it works out. So log base 2 of 8. 2 to what power is equal to 8? Well 2 to the third power
is equal to 8, right? So this term right here,
that equals 3, right? Log base 2 of 8 is equal to 3. 2 to what power is equal to 32? Let's see. 2 to the fourth power is 16. 2 to the fifth power is 32. So this right here is 2 to
the-- This is 5, right? And 2 to the what power
is equal to 256? Well if you're a computer
science major, you'll know that immediately. That a byte can have
256 values in it. So it's 2 to the eighth power. But if you don't know that, you
could multiply it out yourself. But this is 8. And I'm not doing it just
because I knew that 3 plus 5 is equal to 8. I'm doing this independently. So this is equal to 8. But it does turn out that
3 plus 5 is equal to 8. This may seem like magic to
you or it may seem obvious. And for those of you who it
might seem a little obvious, you're probably thinking, well
2 to the third times 2 to the fifth is equal to 2 to
the 3 plus 5, right? This is just an exponent rule. What do they call this? The additive exponent
prop-- I don't know. I don't know the
names of things. And that equals 2 to
8, 2 to the eighth. And that's exactly what
we did here, right? On this side, we had 2
the third times 2 to the fifth, essentially. And on this side, you have
them added to each other. And what makes the logarithms
interesting is and why-- It's a little confusing at first. And you can watch the proofs
if you really want kind of a rigorous-- my proofs
aren't rigorous. But if you want kind of
a better explanation of how this works. But this should hopefully give
you an tuition for why this property holds, right? Because when you multiply
two numbers of the same base, right? Two exponential expressions
of the same base, you can add their exponents. Similarly, when you have the
log of two numbers multiplied by each other, that's
equivalent to the log of each of the numbers added
to each other. This is the same property. If you don't believe me,
watch the proof videos. So let's do a-- Let me show
you another log property. It's pretty much the same one. I almost view them the same. So this is log base b of
a minus log base b of c is equal to log base b
of-- well I ran out. I'm running out of space
--a divided by c. That says a divided by c. And we can, once again, try
it out with some numbers. I use 2 a lot just because
2 is an easy number to figure out the powers. But let's use a
different number. Let's say log base 3 of-- I
don't know --log base 3 of-- well you know, let's make it
interesting --log base 3 of 1/9 minus log base 3 of 81. So this property tells us--
This is the same thing as-- Well I'm ending up
with a big number. Log base 3 of 1/9
divided by 81. So that's the same thing
as 1/9 times 1/81. I used two large numbers
for my example, but we'll move forward. So let's see. 9 times 8 is 720, right? 9 times-- Right. 9 times 8 is 720. So this is 1/729. So this is log base
3 over 1/729. So what-- What does-- 3 to
what power is equal to 1/9? Well 3 squared is
equal to 9, right? So 3-- So we know that if 3
squared is equal to 9, then we know that 3 to the negative
2 is equal to 1/9, right? The negative just inverts it. So this is equal to
negative 2, right? And then minus-- 3 to
what power is equal 81? 3 to the third power is 27. So 3 to the fourth power. So we have minus 2 minus 4 is
equal to-- Well, we could do it a couple of ways. Minus 2 minus 4 is
equal to minus 6. And now we just have to confirm
that 3 to the minus sixth power is equal to 1/729. So that's my question. Is 3 to the minus sixth power,
is that equal to 7-- 1/729? Well that's the same thing as
saying 3 to sixth power is equal to 729, because that's
all the negative exponent does is inverts it. Let's see. We could multiply that out,
but that should be the case. Because, well, we
could look here. But let's see. 3 to the third power-- This
would be 3 to the third power times 3 to the third power
is equal to 27 times 27. That looks pretty close. You can confirm it with
a calculator if you don't believe me. Anyway, that's all the time
I have in this video. In the next video, I'll
introduce you to the last two logarithm properties. And, if we have time, maybe
I'll do examples with the leftover time. I'll see you soon.