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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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Using the properties of logarithms: multiple steps
CCSS.Math:
Sal rewrites log_5([25^x]/y) as 2x-log_5(y) by using both the log subtraction property and the log multiplied by a constant property. Created by Sal Khan and Monterey Institute for Technology and Education.
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What do we do if there are three parts to subtract? All three logs have the same base.(10 votes)
then you just divide divide it 2 times...for example:
log(3/2)/3... what you do: log3 - log 2 - log 3(18 votes)
At0:20can we move the x out in front first, instead of applying the quotient log property first? And then apply the quotient log property?
If not...is there a certain order in which log properties should be applied (like BEDMAS)??? Thanks in advance :)(7 votes)
You can't move the x in front first, because not the whole factor is to the xth power. if we had "log(25/y)^x" we could. otherwise the whole term, so both log25 and log5 would be times 5, which is wrong. I hope this helps you, English is not my first language, so I don't know if I explained it well enough!(18 votes)
- hi this is question i am struggling with any tips or help would be appreciated.
log_10(log_10 x) -log_10(log_10 3) = log_10 (2)
this example states simplify and hence solve for X ,(6 votes)- We start with
log( log(x) ) - log( log(3) ) = log(2). Well, first you can use the property from this video to convert the left side, to getlog( log(x) / log(3) ) = log(2). Then replace both side with 10 raised to the power of each side, to getlog(x)/log(3) = 2. Then multiply through bylog(3)to getlog(x) = 2*log(3). Then use the multiplication property from the prior video to convert the right side to getlog(x) = log(3^2). Then replace both sides with 10 raised to the power of each side again, to getx = 3^2 = 9. And we are done. (And plugging that back in to the original equation with a calculator, it checks out.)
Yo dog, we heard you like logs, so we put logs in your logs so you can log while you log.(17 votes)
If log (a+c) = Log (b+c) does it follow that a+c = b+c ?(5 votes)
Yes, of course. More specifically, it follows that a = b(10 votes)
Who thought of exponents and logarithms?(4 votes)- John Napier (a Scottish mathematician) and Joost Burgi (Swiss mathematician) both discovered them independently around 1600. Napier published "Description of the Marvelous Canon of Logarithms" in 1614, and Burgi published the first lookup tables in 1620.
Source: The Story of Mathematics, A Rooney, p40(7 votes)
- Why is it 2x i got kind of confused sorry?(4 votes)
Log_5 25 = 2. So,x*Log_5 25 = x*2 = 2x(7 votes)
- In my math textbook the answer to a logarithm is given as log_a(sqrt of a/x)
I reasoned that this would equal log_a(sqrt of a) - log_a(x)
The square root of a is equal to a to the 1/2 power. So log_a(a^1/2)=1/2
Therefore, it simplifies to 1/2- log_a(x)
Is this wrong or would the book answer be considered simpler?(3 votes)- Your version is correct and equal to what the textbook said.
You are now moving into a level of math where "simplest form" is not going to be as important as whatever might be a useful form or a convenient form. In fact, some values can be expressed in such a variety of ways that even deciding what "simplest form" might be is subjective.(5 votes)
- How do you find the value of a logarithm if the number is to the power of log? for example, a^8log base a of square root 2 ?(2 votes)
- If the base of both the log and the exponent are the same, they cancel each other out. So loga(a^anything) = anything
It works in the other direction too, so
a^(loga(anything) = anything
As long as "anything" is not 0.
So, in your case you get this:
a^8loga(√2)
= a^loga(√2)^8
(√2)^8 = 2^(8/2) = 2^4 = 16
Let me know it you didn't follow any of that.(4 votes)
I'm just confuse how my teacher arrive on this solution.... So the problem is
3^(x-6) = 22
And he did this
Log of 3 (22) = x-6
(Log of 3 (22)) -6=x
X=-3.19(3 votes)- If you presented your teacher's solution correctly, then your teacher is wrong. Specifically, there is a sign error in dealing with the 6.
Here is one correct way to find the real solution (there are also nonreal solutions, but that is beyond the scope of this level of study).
log₃[3^(x-6)] = log₃ (22)
(x - 6) log₃ 3 = log₃ (22)
(x - 6) = log₃ (22)
x = 6 + log₃ (22)
x≈ 8.81359
Note: if you don't have a calculator that can handle alternative bases of logs, you can use the change of base property: log₃ (22) = ln (22)/ln(3)(2 votes)
- How will you solve 'log to base 2 times x = log to base x times 2'?(2 votes)
0:20Video transcript
We're asked to simplify log base
5 of 25 to the x power over y. So we can use some
logarithm properties. And I do agree that this does
require some simplification over here, that having
this right over here inside of the logarithm is not
a pleasant thing to look at. So the first thing that
we realize-- and this is one of our
logarithm properties-- is logarithm for a given
base-- so let's say that the base is
x-- of a/b, that is equal to log base x of
a minus log base x of b. And here we have
25 to the x over y. So we can simplify. So let me write this down. I'll do this in blue. Log base 5 of 25 to the x over
y using this property means that it's the same thing
as log base 5 of 25 to the x power minus
log base 5 of y. Now, this looks like we can do
a little bit of simplifying. It seems like the relevant
logarithm property here is if I have log base x of
a to the b power, that's the same thing as
b times log base x of a, that this
exponent over here can be moved out front, which
is what we did it right over there. So this part right
over here can be rewritten as x times the
logarithm base 5 of 25. And then, of course, we
have minus log base 5 of y. And this is useful
because log base 5 of 25 is actually fairly
easy to think about. This part right
here is asking us, what power do I have to
raise 5 to to get to 25? So we have to raise 5 to the
second power to get to 25. So this simplifies to 2. So then we are left
with, this is equal to-- and I'll write it in front
of the x now-- 2 times x minus log base 5 of y. And we're done.