Introduction to logarithm properties (part 2) | Logarithms | Algebra II | Khan Academy

PROFESSOR: Welcome back. I’m going to show you the last
two logarithm properties now. So this one– and I always
found this one to be in some ways the most obvious one. But don’t feel bad if
it’s not obvious. Maybe it’ll take a little
bit of introspection. And I encourage you to really
experiment with all these logarithm properties, because
that’s the only way that you’ll really learn them. And the point of math isn’t
just to pass the next exam, or to get an A on the next exam. The point of math is to
understand math so you can actually apply it in life later
on and not have to relearn everything every time. So the next logarithm property
is, if I have A times the logarithm base B of C, if I
have A times this whole thing, that that equals logarithm
base B of C to the A power. Fascinating. So let’s see if this works out. So let’s say if I have 3
times logarithm base 2 of 8. So this property tells us that
this is going to be the same thing as logarithm base 2
of 8 to the third power. And that’s the same thing. Well, that’s the same thing
as– we could figure it out. So let’s see what this is. 3 times log base–
what’s log base 2 of 8? The reason why I kind of
hesitated a second ago is because every time I want to
figure something out, I implicitly want to use log and
exponential rules to do it. So I’m trying to avoid that. Anyway, going back. What is this? 2 to the what power is 8? 2 to the third
power is 8, right? So that’s 3. We have this 3 here,
so 3 times 3. So this thing right
here should equal 9. If this equals 9, then we know
that this property works at least for this example. You don’t know if it works for
all examples, and for that maybe you’d want to look at
the proof we have in the other videos. But that’s kind of a
more advanced topic. But the important thing
first is just to understand how to use it. Let’s see, what is 2
to the ninth power? Well it’s going to be
some large number. Actually, I know what
it is– it’s 256. Because in the last video we
figured out that 2 to the eighth was equal to 256. So 2 to the ninth
should be 512. So 2 to the ninth
should be 512. So if 8 to the third is also
512 then we are correct, right? Because log base 2 of 512
is going to be equal to 9. What’s 8 to the third? It’s 64– right. 8 squared is 64, so 8
cubed– let’s see. 4 times 2 is 3. 6 times 8– looks
like it’s 512. Correct. And there’s other ways
you could have done it. Because you could have said
8 to the third is the same thing as 2 to the ninth. How do we know that? Well, 8 to the third is
equal to 2 to the third to the third, right? I just rewrote 8. And we know from our exponent
rules that 2 to the third to the third is the same
thing as 2 to the ninth. And actually it’s this exponent
property, where you can multiply– when you take
something to exponent and then take that to an exponent, and
you can essentially just multiply the exponents– that’s
the exponent property that actually leads to this
logarithm property. But I’m not going to dwell
on that too much in this presentation. There’s a whole video on
kind of proving it a little bit more formally. The next logarithm property I’m
going to show you– and then I’ll review everything and
maybe do some examples. This is probably the single
most useful logarithm property if you are a calculator addict. And I’ll show you why. So let’s say I have log base B
of A is equal to log base C of A divided by log base C of B. Now why is this a useful
property if you are calculator addict? Well, let’s say you go
class, and there’s a quiz. The teacher says, you can use
your calculator, and using your calculator I want you to figure
out the log base 17 of 357. And you will scramble and look
for the log base 17 button on your calculator,
and not find it. Because there is no log base
17 button on your calculator. You’ll probably either have
a log button or you’ll have an ln button. And just so you know, the log
button on your calculator is probably base 10. And your ln button on
your calculator is going to be base e. For those you who aren’t
familiar with e, don’t worry about it, but it’s 2.71
something something. It’s a number. It’s nothing– it’s an amazing
number, but we’ll talk more about that in a
future presentation. But so there’s only two bases
you have on your calculator. So if you want to figure out
another base logarithm, you use this property. So if you’re given this on an
exam, you can very confidently say, oh, well that is just the
same thing as– you’d have to switch to your yellow color in
order to act with confidence– log base– we could
do either e or 10. We could say that’s the same
thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just
type in 357 in your calculator and press the log button and
you’re going to get bada bada bam. Then you can clear it, or if
you know how to use the parentheses on your calculator,
you could do that. But then you type 17 on your
calculator, press the log button, go to bada bada bam. And then you just divide them,
and you get your answer. So this is a super
useful property for calculator addicts. And once again, I’m not going
to go into a lot of depth. This one, to me it’s the most
useful, but it doesn’t completely– it does fall
out of, obviously, the exponent properties. But it’s hard for me to
describe the intuition simply, so you probably want to watch
the proof on it, if you don’t believe why this happens. But anyway, with all of those
aside, and this is probably the one you’re going to be using
the most in everyday life. I still use this in my job. Just so you know
logarithms are useful. Let’s do some examples. Let’s just let’s just
rewrite a bunch of things in simpler forms. So if I wanted to rewrite the
log base 2 of the square root of– let me think of something. Of 32 divided by the cube– no,
I’ll just take the square root. Divided by the
square root of 8. How can I rewrite this so
it’s reasonably not messy? Well let’s think about this. This is the same thing, this
is equal to– I don’t know if I’ll move vertically
or horizontally. I’ll move vertically. This is the same thing as
the log base 2 of 32 over the square root of 8 to
the 1/2 power, right? And we know from our logarithm
properties, the third one we learned, that that is the same
thing as 1/2 times the logarithm of 32 divided by the
square root of 8, right? I just took the exponent and
made that the coefficient on the entire thing. And we learned that in the
beginning of this video. And now we have a little
quotient here, right? Logarithm of 32 divided by
logarithm of square root of 8. Well, we can use our
other logarithm– let’s keep the 1/2 out. That’s going to equal,
parentheses, logarithm– oh, I forgot my base. Logarithm base 2 of
32 minus, right? Because this is in a quotient. Minus the logarithm base 2
of the square root of 8. Right? Let’s see. Well here once again we have a
square root here, so we could say this is equal to 1/2
times log base 2 of 32. Minus this 8 to the 1/2,
which is the same thing is 1/2 log base 2 of 8. We learned that property in the
beginning of this presentation. And then if we want, we can
distribute this original 1/2. This equals 1/2 log base 2 of
32 minus 1/4– because we have to distribute that 1/2–
minus 1/4 log base 2 of 8. This is 5/2 minus, this is 3. 3 times 1/4 minus 3/4. Or 10/4 minus 3/4
is equal to 7/4. I probably made some arithmetic
errors, but you get the point. See you soon!

100 thoughts on “Introduction to logarithm properties (part 2) | Logarithms | Algebra II | Khan Academy

  1. @goldensilverstar Can't tell, but I have to use it to calculate frequency table in my classes of probability at the university. I'm studying Computer Science, and I forgot some principles…cough! … to be blunt, I forgot everything about logarithms. There's some rumors I'll use it also in circuit classes. This I'll see, soon.

  2. wait how is log base2 (sqrt32/sqrt8) = log base2 (32/sqrt8)^(1/2)? shouldn't it only be 32^(1/2)? like (32^(1/2)/sqrt8) isnt that right? wtf?

  3. I often find myself sitting in class watching what's being written on the board and think " when will i ever need this?!"
    after watching 2 of your videos i can actually see myself using logarithms in a real life situation and not just on a test to get a grade for school 😀
    thanks man 😉

  4. @4:15…. Lol I do that all the time, I will say what I'm going to write, and then just sit there stroking the pencil in mid-air… I hate when I do that….. Lol He does it twice at 8:27 too!

  5. I was not a fan of government-funded cloning programs . . . until Khan showed up. *cue Imperial Theme*

  6. I didn't get the last part clearly. The way the problem is written at the beginning of the last example is easy to enter into EXCEL. "=LOG(SQRT(32/SQRT(8)),2)". Other then getting an A for an exam, why would one need to know all the rules you used to solve this problem?

  7. Haha! I love the confidence inference:D this made learning fun. I skipped a week of school and thank heavens for khan academy(:

  8. I wish i had realized that years ago. But its not too late to apply and try to understand it i guess. Being a info systems major, I think im required to understand it lol

  9. I cant take it anymore. You are actually quite effective in communicating the material. however, YOU DIGRESS (A LOT) FROM THE POINT/SUBJECT so much that it is kind of confusing. Also to add to the confusion is the "machine" you are using. However, thanks for the free material it did help a lot!

  10. Thanks a bunch for these tutorials, they were the best I could find. And since I missed my theory lessons of logarithms, this was very useful! I now begin to understand it better. Cheers!

  11. To do log with a different base on a graphing calculator hit the math button then scroll down till you see logBASE( to give the log a different base

  12. I just do NOT understand why he kept the 1/2 out of parentheses. This would have never intuitively occurred to me if I were doing this. It seems to be completely arbitrary.

  13. I'll explain how he got to convert log base 2 (32) and log base 2 (8). long base 2 (32) is 2 to the X equals 32. 32 can be divided into 8 times 4. Each of these can be factored into 2 to the 3rd and 2 to the second respectively. That will make log base 2 (32) equal to 5. then you multiply 5 with the coefficient 1/2 to get 5/2. Similarly, log base 2 (8) is equal to 3, then multiply it with 1/4 to get 3/4. You add 1/2 to 3/4 and you get 7/4.

  14. This was good but the last question couldve been solved easily by taking 32 as 2^5 and 8 as 2^3 and then multiplying them with 1/2 coz of the sq root….or maybe thats just coz im an Indian 😛

  15. after getting 9 in the first problem, I got lost with the process of getting the final answer, 512, why did we do 2 ^9?

  16. "the purpose of math isn't just to pass your next test, but to understand math" No, passing my test is the only reason I'm learning this. I missed the entire unit of Logarithms because I was sick, and here I am 2 months after that unit, trying to learn what I missed so I can take the bloody test that the rest of my class took 2 months ago, so I don't fail.

  17. i have three bases on my calculator. The two you mentioned in the video and I also have an option to pick which base I like.

  18. At 8:15 wouldn't the root 8 just become 8 when you put it to the power of a half. Same as the root 2 became 32? Log base 2 of (root 32 /root 8) = 1. Log base 2 of (32/root 8)1/2 power = 1.75. Correct me if I'm wrong. Can't see how you can cancel the root on the top half of the fraction and put 32 to the power of a half but leave the bottom half with the root and still put it to the power of a half

  19. Where's the video that explains more formally the multiplication of logarithms? He says it right at about 3:55. where's that video?

  20. If you don’t believe why this is true you can go watch the proof. Nope you could say pigs fly and I wouldn’t question it

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