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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- 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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Intro to logarithm properties
Learn about the properties of logarithms and how to use them to rewrite logarithmic expressions. For example, expand log₂(3a).
The product rule | log, start base, b, end base, left parenthesis, M, N, right parenthesis, equals, log, start base, b, end base, left parenthesis, M, right parenthesis, plus, log, start base, b, end base, left parenthesis, N, right parenthesis | |
The quotient rule | log, start base, b, end base, left parenthesis, start fraction, M, divided by, N, end fraction, right parenthesis, equals, log, start base, b, end base, left parenthesis, M, right parenthesis, minus, log, start base, b, end base, left parenthesis, N, right parenthesis | |
The power rule | log, start base, b, end base, left parenthesis, M, start superscript, p, end superscript, right parenthesis, equals, p, dot, log, start base, b, end base, left parenthesis, M, right parenthesis |
(These properties apply for any values of M, N, and b for which each logarithm is defined, which is M, N, is greater than, 0 and 0, is less than, b, does not equal, 1.)
What you should be familiar with before taking this lesson
You should know what logarithms are. If you don't, please check out our intro to logarithms.
What you will learn in this lesson
Logarithms, like exponents, have many helpful properties that can be used to simplify logarithmic expressions and solve logarithmic equations. This article explores three of those properties.
Let's take a look at each property individually.
The product rule: log, start base, b, end base, left parenthesis, M, N, right parenthesis, equals, log, start base, b, end base, left parenthesis, M, right parenthesis, plus, log, start base, b, end base, left parenthesis, N, right parenthesis
This property says that the logarithm of a product is the sum of the logs of its factors.
We can use the product rule to rewrite logarithmic expressions.
Example: Expanding logarithms using the product rule
For our purposes, expanding a logarithm means writing it as the sum of two logarithms or more.
Let's expand log, start base, 6, end base, left parenthesis, 5, y, right parenthesis.
Notice that the two factors of the argument of the logarithm are start color #11accd, 5, end color #11accd and start color #1fab54, y, end color #1fab54. We can directly apply the product rule to expand the log.
Example: Condensing logarithms using the product rule
For our purposes, compressing a sum of two or more logarithms means writing it as a single logarithm.
Let's condense log, start base, 3, end base, left parenthesis, 10, right parenthesis, plus, log, start base, 3, end base, left parenthesis, x, right parenthesis.
Since the two logarithms have the same base (base-3), we can apply the product rule in the reverse direction:
An important note
When we compress logarithmic expressions using the product rule, the bases of all the logarithms in the expression must be the same.
For example, we cannot use the product rule to simplify something like log, start base, 2, end base, left parenthesis, 8, right parenthesis, plus, log, start base, 3, end base, left parenthesis, y, right parenthesis.
Check your understanding
The quotient rule: log, start base, b, end base, left parenthesis, start fraction, M, divided by, N, end fraction, right parenthesis, equals, log, start base, b, end base, left parenthesis, M, right parenthesis, minus, log, start base, b, end base, left parenthesis, N, right parenthesis
This property says that the log of a quotient is the difference of the logs of the dividend and the divisor.
Now let's use the quotient rule to rewrite logarithmic expressions.
Example: Expanding logarithms using the quotient rule
Let's expand log, start base, 7, end base, left parenthesis, start fraction, a, divided by, 2, end fraction, right parenthesis, writing it as the difference of two logarithms by directly applying the quotient rule.
Example: Condensing logarithms using the quotient rule
Let's condense log, start base, 4, end base, left parenthesis, x, cubed, right parenthesis, minus, log, start base, 4, end base, left parenthesis, y, right parenthesis.
Since the two logarithms have the same base (base-4), we can apply the quotient rule in the reverse direction:
An important note
When we compress logarithmic expressions using the quotient rule, the bases of all logarithms in the expression must be the same.
For example, we cannot use the quotient rule to simplify something like log, start base, 2, end base, left parenthesis, 8, right parenthesis, minus, log, start base, 3, end base, left parenthesis, y, right parenthesis.
Check your understanding
The power rule: log, start base, b, end base, left parenthesis, M, start superscript, p, end superscript, right parenthesis, equals, p, log, start base, b, end base, left parenthesis, M, right parenthesis
This property says that the log of a power is the exponent times the logarithm of the base of the power.
Now let's use the power rule to rewrite log expressions.
Example: Expanding logarithms using the power rule
For our purposes in this section, expanding a single logarithm means writing it as a multiple of another logarithm.
Let's use the power rule to expand log, start base, 2, end base, left parenthesis, x, cubed, right parenthesis.
Example: Condensing logarithms using the power rule
For our purposes in this section, condensing a multiple of a logarithm means writing it as another single logarithm.
Let's use the power rule to condense 4, log, start base, 5, end base, left parenthesis, 2, right parenthesis,
When we condense a logarithmic expression using the power rule, we make any multipliers into powers.
Check your understanding
Challenge problems
To solve these next problems, you will have to apply several properties in each case. Give it a try!
Want to join the conversation?
- Why isn't there a base in question 4?(26 votes)
- Since both of them have no base you can presume they have the same base (no base) so you can apply the property of log. Your answer would be without base as well.
Later on you'll learn that no base log is understood as base of 10.(100 votes)
- in the 1st challenging problem the ans given is (c) instead of a.if simplified
we get log base(b) (2x^3/5)= 3log base(b) 2+3log base(b) x - log base(b) 5. it shud be either a(9 votes)- Your mistake is in dealing with log_b (2x^3).
It is only the x that is cubed, not the 2.
So we first need to separate the factors (2 and x^3), getting
log_b (2) + log_b (x^3).
Only then can we deal with the cube, getting
log_b (2) + 3 log_b (x)(45 votes)
- why the base cant be 1 of a logarithm?(6 votes)
- Let take an example of log₁(25)=x
This would mean 1^x=25
But 1 to any powers will always be 1.
So the base can't be 1 because it would make the log expression false, unless log₁(1)=x but then x would be any and all real numbers. So the convention is to rule out log base 1.(45 votes)
- Can the answer to logarithm be negative?(4 votes)
- Any number less than 1 but greater than 0 will have a negative logarithm*
eg ln(0.1) = -2.303(17 votes)
- Is there a distributive property for logarithms? For example, does ln(a+b)=ln(a)+ln(b)? I don't think I would use the product rule here. Is this covered in another lesson? Thanks!(3 votes)
- Properties like this are covered later in the logarithms playlist. The property you suggest doesn't hold. ln(1+2)=ln(3), but ln(1)+ln(2)=0+ln(2)=ln(2), and ln(3)≠ln(2).
The relevant property here is that ln(ab)=ln(a)+ln(b). If you start from the property e^xe^y=e^(x+y), you can take the natural log of both sides to get
ln(e^xe^y)=x+y
Now let e^x=a (or x=ln(a)) and e^y=b (or y=ln(b)) and substitute in to get ln(ab)=ln(a)+ln(b).(8 votes)
- It looks like the answer to challenge question is wrong if "C" is answer.
Should be 3log_B(2) + 3log_B(X) - log_B(5).
Answer "C" leaves out the multiplier "3" in the first term of the solution.
Right ?(0 votes)- The three exponent only refers to the x, ao you can't expand using the power rule until you first use the product rule to get rid of the 2: log_b(2) + log_b(x^3) - log_b(5) and then expand the x^3 to get log_b(2) + 3log_b(x) - log_b(5).(9 votes)
- why logarithms used in graphs ?(3 votes)
- Interesting question as a fraction of people might get confused in it.
Log is the inverse function of exponent which you can say another form of exponent...so like we use exponents in graph eg ---> y = x^2 same we can also write log_x (y) = 2
And both are the same but at some point things get easier while using log.
A bit of a practice and experience and you will understand what I am saying.
Have a good day !!(3 votes)
- where is the base(2 votes)
- logₐ(b) = a is your base
if it's simply just log(b), always assume 10 is your base
if it's ln(b), always assume that "e" (an irrational constant like pi) is the base.
hopefully that helps !(5 votes)
- When condensing and the original problem begins with ln, do you need to switch it to log base e or can you leave it as ln?
ex: 2 ln 5 - 3 ln 2
final answer: log base e (25/8) or ln (25/8)? thank you :)(3 votes)- "Log base e" and "ln" are the same. You can use either one, though using "ln" is less to write.(2 votes)
- If 2log(x+4)-log(x+12)=log10, I get x=-8,4.
Do I have to eliminate -8 from the answer because the -8 makes (x+4) a negative number? Or I keep -8 because the coefficient attached to the log (x+4) turns (-4)^2=16?(3 votes)- You are correct that when x = -8, the expression (x + 4) becomes negative, which is not valid because the logarithm of a negative number is undefined. However, you should keep both answers (-8 and 4) and check them separately to see which one satisfies the original equation.
When x = 4, 2log(x+4) - log(x+12) = 2log(8) - log(16) = 2(3) - 1 = 5, and log(10) = 1, so the left-hand side of the equation matches the right-hand side. Therefore, x = 4 is a valid solution.
When x = -8, 2log(x+4) - log(x+12) = 2log(-4) - log(4) = undefined, so x = -8 is not a valid solution.
Therefore, the only solution to the equation is x = 4.(2 votes)