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## Algebra 2

### Course: Algebra 2 > Unit 8

Lesson 3: Properties of logarithms- 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

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# Using the logarithmic product rule

CCSS.Math:

Sal rewrites log₃(27x) as log₃(27)+log_3(x), which is simplified as 3+log₃(x). Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- is there a way to simplify log3^x?(14 votes)
- Are you referring to log_3^x or are you asking about Log_10 3^x?

Log base 3 of x is simplified.

However Log base 10 3^x can be further simplified by using the Power Property which states in any logarithm with a standard base of 10 or e a constant raised to a power can be broken down like so:

Log_10 3^x the x can be taken and moved to the front of the equation to render xLog_10 3.

Hopefully this helped and is an accurate answer.(43 votes)

- So is there any way to
*solve*for x?(6 votes)- Not in this case, because it is an expression not an equation. For example can you solve for "x" here?

10x

What is x? It could be anything. If we had an equation on the other hand:

10x = 4

Then we could solve for x.(14 votes)

- :( I have been way too confused with logarithms and I don't know why. It seems so easy to understand in my head, but when I try it out myself I don't know what to do. I'm really confused with what the answer is in the end.

Is it just a number or do you end with a logarithm? And in the last problem I don't get how that was the answer?(5 votes)- I think Sal was just demonstrating the properties of logarithms. I think if you plug in those alphabets with small numbers like 2, you will start to understand it. And if you go all the way to the end, you'll get an actual answer, or one that makes sense.(2 votes)

- Express as a sum or difference of logarithms without exponents.

log(base b7)Square root (x^6)/y^7z^8

What is the equivalent sum or difference of logarithms?(4 votes)- I am very good with logs, but I need you to clarify what you mean.

Did you mean log₇{(√x⁶) / (y⁷z⁸)} ?

If so you would solve it as follows:

log₇{(√x⁶) / (y⁷z⁸)}

= log₇(√x⁶) - log₇(y⁷) - log₇(z⁸)

=log₇(x³) - log₇(y⁷) - log₇(z⁸)

= 3log₇(x) - 7log₇(y) - 8log₇(z)(4 votes)

- in the video sum of logarithm with same base how do you find x(3 votes)
- Since there's no equality, we cannot find x. He is just simplifying a statement.(4 votes)

- If we have Log_3(27(y^3)) = log_3(3^3 * y^3), shouldn't it turn out like this = 3+Log_3(3y)?(3 votes)
- You're super close. Since we have an exponent in the logarithm, when we take it out it becomes a number multiplied to the log. The solution would look like this:

log_3(27y^3)

log_3((3y)^3)

3 * log_3(3y)(3 votes)

- Are the rules and stuff still the same if you are given a
*minus*in the equation instead of*addition*?

Ex: log(10)-log(2) rewrite in log(c) form(2 votes)- The minus indicated division: Log(10/2)

A plus would be multiplication.(2 votes)

- At4:39

log₃(27x) as log₃(27)+log_3(x), which is simplified as 3+log₃(x)

I mean why is that? cause log₃(27) suppose to be log₃(3^3), then =3 log₃3(1 vote)- Your work is ok so far, but is incomplere as
`3 log₃3`

can be simplified: `3 log₃3 = 3*1 = 3.

Remember, 3^1 = 3. So, log₃3 = 1.

Hope this helps.(3 votes)

- Why do the same bases cancel out?(2 votes)
- Could you be more specific please....But if you mean Simplify log_b(b^3).

The Relationship says that "log_b(b^3) = y" means "b ^y = b^3". Then clearly y = 3, so:

log_b(b^3) = 3

This is always true: log_b(b^n) = n for any base b.

Some students like to think of the above simplification as meaning that the b and the log-base-b "cancel out". This is not technically correct, but it can be a useful way of thinking of things. Just don't say it out loud in front of your instructor.(1 vote)

- I was sure I learned a lot about maths thanks to khan, but when I tried to do an admission test for my university, I realized I couldn't answer to any of the questions. What do you think I have to study in order to be admitted?(1 vote)
- I know this is late:

Calculus I - III: Professor Leonard on YouTube

Linear Algebra: Gilbert Strang on MIT OCW

Specific problems/concepts: PatrickJMT(3 votes)

## Video transcript

We're asked to simplify
log base 3 of 27x. And frankly, this is
already quite simple. But I'm assuming they want us
to use some logarithm properties and manipulate this in
some way, maybe to actually make it a little bit
more complicated. But let's give our
best shot at it. So the logarithm property
that jumps out at me-- because this right over here--
we're saying what power do we have to raise 3 to to get 27x? 27x is the same
thing as 27 times x. And so the logarithm property it
seems like they want us to use is log base-- let me write
it-- log base b of a times c-- I'll write it this way--
log base b of a times c. This is equal to the
logarithm base b of a plus the logarithm base b of c. And this comes straight out
of the exponent properties that if you have two exponents,
two with the same base, you can add the exponents. So let me make that a
little bit clearer to you. And if this part is
a little confusing, the important part
for this example is that you know
how to apply this. But it's even better if
you know the intuition. So let's say that log base b
of a times c is equal to x. So this thing right over
here evaluates to x. Let's say that this thing
right over here evaluates to y. So log base b of
a is equal to y. And let's say that this thing
over here evaluates to z. So log base b of
c is equal to z. Now, what we know is, this thing
right over here or this thing right over here tells
us that b to the x power is equal to a times c. Now, this right over
here is telling us that b to the y
power is equal to a. And this over here is telling
us that b to the z power is equal to c. Let me do that in
that same green. So I'm just writing
the same truth. I'm writing it as an
exponential function or exponential equation, instead
of a logarithmic equation. So b to the zth
power is equal to c. This is the same statement
or the same truth said in a different way. And this is the same truth
said in a different way. Well, if we know that
a is equal to this, is equal to b to the y,
and c is equal to bz, then we can write
b to the x power is equal to b to the y
power-- that's what a is. We know that already--
times b to the z power. And we know from our
exponent properties that if we take b to
the y times b to the z, this is the same thing
as b to the-- I'll do it in a neutral color--
b to the y plus z power. This comes straight out of
our exponent properties. And so if b to
the y plus z power is the same thing as b to
the x power, that tells us that x must be
equal to y plus z. If this is confusing to you,
don't worry about it too much. The important thing, or at
least the first important thing, is that you know
how to apply it. And then you can think about
this a little bit more, and you can even try it
out with some numbers. You just have to realize
that logarithms are really just exponents. I know when people first would
tell me that, I was like, well, what does that mean? But when you
evaluate a logarithm, you're getting an
exponent that you would have to raise b
to to get to a times c. But let's just apply this
property right over here. So if we apply it
to this one, we know that log base
3 of 27 times x-- I'll write it that way-- is
equal to log base 3 of 27 plus log base 3 of x. And then this right over
here, we can evaluate. This tells us, what
power do I have to raise 3 to to get to 27? You could view it as this way. 3 to the ? is equal to 27. Well, 3 to the third
power is equal to 27. 3 times 3 is 9 times 3 is 27. So this right over
here evaluates to 3. So if we were to
simplify-- or I guess I wouldn't even call
it simplifying it. I would just call it expanding
it out or using this property, because we now have two
terms where we started off with one term. Actually, if we
started with this, I'd say that this is the
more simple version of it. But when we rewrite it,
this first term becomes 3. And then we're left with
plus log base 3 of x. So this is just an
alternate way of writing this original statement,
log base 3 of 27x. So once again, not
clear that this is simpler than this
right over here. It's just another way of writing
it using logarithm properties.