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## Algebra 2 (Eureka Math/EngageNY)

### Course: Algebra 2 (Eureka Math/EngageNY)>Unit 3

Lesson 5: Topic B: Lessons 10-12: Logarithm properties

# Using the logarithmic product rule

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?
• 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.
• So is there any way to solve for x?
• 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.
• :( 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?
• 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.
• 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?
• 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)
• in the video sum of logarithm with same base how do you find x
• Since there's no equality, we cannot find x. He is just simplifying a statement.
• 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)?
• 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)
• 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
• The minus indicated division: Log(10/2)
A plus would be multiplication.
• At
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.
• Why do the same bases cancel out?
• 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)
• so in he said we just add the exponents together to simplify them. this is the part I'm confused on right now.
(1 vote)
• the b^y*b^z=b^y+z is a rule that you would learn when learning about exponents.
for example,
2^3*2^4 would become 2^7, because (2*2*2)*(2*2*2*2) is 2*7.
note: only works when the b is the same value for both equations. you can't do it with, say 3^2 and 2^3. it doesn't become 2^5 or 3^5.