Question

Note: This is NOT a question. This is a tutorial.

What are the fundamental properties of logarithms? See comment below to see what!

For beginners, see my tutorial. For more advanced people, scroll down through the comments to see the discussion of KingGeorge, satellite73, and amistre64 on other thingies that can be done with logarithms

Answer 1

First of all, we need to understand the definition of logarithms.

Definition of logarithms:

\(\large \log_{a} b = x\)

this means that x is the exponent that a must be raised to obtain b. Exponentially, we can rewrite it as

\(\large a^{x} = b\)

Now, on to the fundamental properties...

Property 1: \(\large \log_{a} a = 1\) why? Simply because 1 is the only exponent you can raise a that it'll come out as itself.

Example: \(\log_{3} 3 = 1\)

Property 2: \(\large \log_{a} 1 = 0\) why? Because anything raised to zero is one. Of course, this property is only true if a is not equal to 1 nor zero. Why? Because if a is 1, there would be an infinite number of possible answers as 1 raised to anything will equal 1. On the other hand, \(0^0\) would not wield 1 either. \(0^0\) is an indeterminate form because it is a "gray area" between the properties "anything raised to zero equals one" and "zero raised to anything is zero". It satisfies both conditions, but it cannot be determined which is the answer for \(0^0\)

Example: \(\log_{2} 1 = 0\)

Property 3: \(\log_{a} a^x = x\) or \(a^{log_{a} x} = x\) These two are related. If we read the former, it is "what can we raise a to obtain \(a^x\)". Which obviously, the answer is x. For the latter, if we transform it to exponential form (which i am assuming you know how), you'll get \(\log_{a} x = \log_{a} x\) which shows that it is merely an identity property, the same with the former.

Example: \(\log_{2} 2^3 = 3\)

\(3^{log_{3} 2} = 2\)

Product Law: \(\log_{a} x + \log_{a} y = \log_{a} xy\) note that the two addends MUST have THE SAME base. It merely states that if you add two logarithms with the same base, it is also equal to getting the product of your two powers (which in this case is x and y), then taking its logarithm to that same base (which in this case is a)

Example: \(\log_{2} 3 + \log_{2} 5 = \log_{2} 15\) <---try inputting that in a scientific calculator and see if you get the same answers. *puke rainbow*

Quotient Law: \(\large \log_{b}c - \log_{b}e = \log_{a} \frac{x}{y}\). This is somehow similar to the Product Law. In this property, it is stated that if it's the difference of two logaritms with the same base, it is also equal to taking the quotient of your two powers (which in this case is c and e) then taking its logarithm to that same base (which in this case is b).

Example: \(\log_{3} 12 - \log_{3} 6 = \log_{3} 2\)<----again, you can try inputting it in the calculator to verify, but I'm sure you've ran out of rainbows to puke :D

Power Law: \(\log_{a} b^x = x\log_{a} b\) Note that your base and power must NOT be the same. Otherwise, it is Property 3. This property merely states that if you have a power (which in this case is \(b^x\)) that also has an exponent, it is equal to solving for the logarithm of the base of the power (the power is \(b^x\) so the base of the power is b) to the base of the original logarithm (in this case, the base of our original logarithm is a) then multiplying it to the exponent of the power(which in this case is x). The definition I stated may sound confusing so a demonstration is in order.

Example: \(\log_{2} 3^4 = 4 \log_{2} 3\) <---you know what to do..bring out those calculators and verify! *pukes pot of gold*

Fun Fact: We can use the Power Rule to prove the first part of Propert 3. \(\log_{a} a^x\). If we apply Power Rule, it will become \(x\log_{a} a\). And what is \(\log_{a} a\)? It's 1! So, x(1) = x, thus proving that \(\log_{a} a^x = x\).

Change-of-Base Formula. \(\Large \log_{a} b = \frac{\log_{c} b}{\log_{c} a}\). This, in my opinion, is the most confusing and the hardest to explain of all the fundamental properties. This property states that if you have a logarithm of b to the base a, it is also the equal to taking the logarithm of b to a certain base, then dividing it by the logarithm of a taken to the same base. You may ask, why is this essential? Why take two logarithms when you can only take 1? This property has many uses. Two of which is to transform bases. For example, you have \(\log_{3} 2\) but you need it as a base of two. You can use this property to change its base to 2. \(\Large \log_{3} 2 = \frac{\log{2} 2}{\log_{2} 3} = \frac{1}{\log_{2} 3}\) Now, it is in the base of 2. Another use? For usage of calculators. You cannot just merely input \(\log_{3} 2\) in a calculator, can you? (Actually some calculators can, but for the sake of explaining I'll asume you don't have that calculator). We can use this formula to do so! \(\log_{3} 2 = \frac{\log 2}{\log 3}\). You will learn later on what those invisible bases mean. For now, all you need to know is that \(\log 2\) and \(\log 3\) have the same "invisible bases", and that those logarithms with "invisible bases" are solveable by the common scientific calculator, and by using this property, you can solve for logarithms faster.

Example: \(\log_{4} 2\) We are going to solve this in two ways. The first is the Change-of-Base Formula, the second is the application of the other fundamental properties to verify the answer.

By C-O-B: \(\Large \log_{4} 2 = \frac{\log_{2} 2}{\log_{2} 4} = \frac{1}{\log_{2} 2^2} = \frac{1}{2}\)

By Fundamental Properties: \(\Large \log_{4} 2 = \log_{4} 4^{\frac{1}{2}} = \frac{1}{2}\) \(\checkmark\)

Answer 2

@KingGeorge

Answer 3

Gj @lgbasallote ,Thats awesome :)

Answer 4

thanks @Eyad :D

Answer 5

Looking good.

Answer 6

i had troubles with the change of base :/

Answer 7

Haha lol, I just learned logarithms a hour before this post, good job :D

Answer 8

well now you know more :P btw Special mention to @zepp for showing me that tutorials can be funny by adding cliched puns :D

Answer 9

http://puu.sh/sXBu
log22 ;)

Answer 10

ARGHHHHHHHHH REWRITE!!!

\LARGE \log_{3} 2 = \frac{\log_{2} 2}{\log_{2} 3}\)

Answer 11

\(\LARGE \log_{3} 2 = \frac{\log_{2} 2}{\log_{2} 3}\)

Answer 12

spose we simply define a set of functions that follow this simple rule:

L(xy) = L(x) + L(y)

and that any function that satisfies this condition is "Logarithmic"

Answer 13

suppose? isn't it?

Answer 14

based on this property; we should be able to define all the other properties that are so "mysteriously" logarithms i believe

Answer 15

L(x) = L(x) ; why?

L(x*1) = L(x) + L(1)
= L(x) + 0

therefore L(1) = 0

Answer 16

L(x^2) = L(x*x)
= L(x) + L(x)
= 2 L(x)

Answer 17

i see...that is cool *_* they never teach you that :/

Answer 18

Logarithms can be considered of without ever knowing what a Logarithm is :) just based on defining functions that have the property L(xy) = L(x) + L(y)

L(x/x) = L(1)
L(x* 1/x) = 0
L(x) + L(1/x) = 0

therefore:
L(1/x) = -L(x)

Answer 19

L(x/y) = L(x* 1/y)
= L(x) + L(1/y)

from above we can rewrite as

L(x/y) = L(x) - L(y)

Answer 20

http://www.cosmolearning.com/video-lectures/logarithms-without-exponents-10784/

im starting to love this guy :) his "hi" is a little creepy, but ehh....

Answer 21

but i dont think it's easy to understand without the basic knowledge...i mean i was amused because it's a fascinating shortcut but i already know the fundamental properties...is it easy to learn withoout that knowledge?

Answer 22

its simple enough to memorize; but if we really want to make any sense of it, we hav to delve deeper into it.

The properties seem almost magical at first and fly in the face of what we are used to knowing as math.

But, by knowing the fundamental basis of a Logarithmic function, and how its properties are defined and made rigorous thru calculus; we gain a deeper appreciation for them and the mystery fades

Answer 23

that was deep...but based on what i comprehended i agree :)`

Answer 24

about 13:15 in that video i linked he starts into what I have shared :)

Answer 25

i wanna see the creepy hi >:))

Answer 26

lol ive watched alot of his videos and they all start out with that creepy "hi" :)

Answer 27

hahaha i keep rewatching it for that hi :p

Answer 28

ive been thinking about how to go about defining a base to this system

\[\frac{L(b)}{L(b)}=1;\ b\ne1\]

as such we can define a base so be the value of 1/L(b)

how do we come up with a logically consistent change of base formula from this?

\[L_b(x)=\frac{L(x)}{L(b)},\ for\ reference \]

spose we want to change the base from L(b) to L(b') . Wouldnt we just multiply by L(b)/L(b')?

\[\frac{L(x)}{L(b)}*\frac{L(b)}{L(b')}=\frac{L(x)/L(b)}{L(b')/L(b)}=\frac{L(x)}{L(b')}\]

\[L_b(x)= \frac{L_{b'}(x)}{L_{b'}(b)}=\frac{L(x)/L(b')}{L(b)/L(b')}=\frac{L(x)}{L(b)}\to\ L_b(x) \]

Answer 29

hmm makes sense...

Answer 30

@lgbasallote, while your at it, write a book :P

Answer 31

am working on it :P hahaha

Answer 32

you can show all these properties belong to
\[L(x)=\int_1^x\frac{1}{t}dt\] directly

Answer 33

really? the properties are in integral? :O

Answer 34

for example, you can show that
\[L(ab)=L(a)+L(b)\] by showing that
\[\int_1^{ax}\frac{1}{t}dt=\int_1^x\frac{1}{t}dt +\int_1^a\frac{1}{t}dt\]

Answer 35

yes they are definitely in the integral because the most precise definition of \(\log(x)\) is
\[\log(x)=\int_1^x\frac{1}{t}dt\]

Answer 36

ommigosh o.O

Answer 37

I'd also like to chime in saying that logarithms are not restricted to the real numbers. In fact, they're commonly used in cyclic groups of prime order as well. Although you only really see this application in number theory, and/or cryptography.

Answer 38

as in \(\mathbb{Z_p}\) we have the "discrete logarithm"

Answer 39

Without which the Diffie-Hellman Key Exchange Protocol would be much more difficult, Along with El-Gamal Public Key Cryptosystem.

Answer 40

and dont forget \(\log(z)=\ln(|z|)+i\theta\)

Answer 41

And one more property

\[\huge a^{\log_b c }= c^{\log_b a}\]

Answer 42

Tell me more about logarithms of complex numbers
\[\log(z)=\ln(|z|)+iθ\]
Where does this formula come from ?
How come the base changes?

Answer 43

@UnkleRhaukus
z can be expressed as\[ z = |z|e^{i(\theta)}\]
Now apply log on both sides and get your required answer

Answer 44

How come the base changes?
--->Actually it should be
\[\ln(z)=\ln(|z|)+i\theta\]

Answer 45

\[~~~~~~z=|z|e^{iθ}\]\[\qquad\downarrow\]\[\{\text{taking the natural logarithm}\}\]\[\qquad\downarrow\]\[\ln(z)=\ln|z|e^{iθ}\]\[=\ln |z|+i\theta\ln e\]\[=\ln|z|+i\theta\]

Answer 46

Exactly @UnkleRhaukus

Answer 47

Is the wikipedia page also incorrect @shivam_bhalla ?

http://en.wikipedia.org/wiki/Complex_logarithm

Answer 48

From my reading of the wiki page, the complex log doesn't have a specified base. The log is defined in the manner that makes the most sense. It's just the component \(\ln(r)\) that has an actual base.

Answer 49

Additionally, this can't be seen as an ordinary logarithm since, the property \[\log(ab)=\log(a)+\log(b)\]fails.

Answer 50

Yes We are specifically considering to the base e while taking log. @KingGeorge is right :)

Answer 51

Practice Question. What real number values of x satisfy the following logarithmic equation? See attachment.

Answer 52

\(\huge (x+1)^{\log_{2}(x+1)^8}=3\)
\(\huge \log_{x+1}3=\log_2 (x+1)^8\)
Let's say x+1 = a

Then \(\huge \log_{a}3=8\log_2 a\)
\(\huge 8 = \frac{log_a 3}{log_2 a}\)

And then I'm stuck :(

Answer 53

@KingGeorge

Answer 54

It's easy
\[8(\log_2(x+1))^2 = 8\]
\[(\log_2(x+1))^2=1\]

Answer 55

I am sorry but I just shortened my steps to a great extent :D

Answer 56

lol where does that square come from? o.o

Answer 57

\[\Large 8 = \frac{log_a 3}{log_2 a}=\frac{\frac{log_2 3}{log_2a}}{log_2 a}\] \[\Large=\frac{log_2 3}{\left(log_2a\right)^2}\]

Answer 58

Simplify a bit more and you get what @shivam_bhalla got.

Answer 59

OK :D. Looks like I should give you a hint

Use this property

\[\log_a (b^c) = c \log_a (b)\]

Answer 60

So \(\huge 8=\frac{log_2 3}{(log_2 a)^2}\)
\(\huge 8(log_2 a)^2 = log_2 3\)
?

Answer 61

8(log2(x+1))2=8 I can't get the 8 you have on the right side D;

Answer 62

LOL. Let me show it to you in another way :D

\[\large \log_2{(x+1)^{\log_2{(x+1)}^8}}=8\]
Now using the property I mentioned before, we get
\[\large(\log_2{(x+1)}^8) (\log_2{(x+1)})=8\]
Now again applying the property,
\[\large8(\log_2{(x+1)}) (\log_2{(x+1)})=8\]
which is
\[\large8(\log_2{(x+1)}^2)=8\]

Answer 63

Sorry Typo in the last step. It should be
\[\large8(\log_2{(x+1)})^2=8\]

Answer 64

OH! I see I see :D Thanks @shivam_bhalla!

And @KingGeorge Isn't \(\log_2{8} =3\) ? \(\log_2(3)\) gives me 1.58496..

Answer 65

I realized that -.- I am disappoint.

Answer 66

@zepp , wasn't it easy. :D Now take two cases
Case 1->
\[\large \log_2{(x+1)}=1\]

Case 2->
\[\large \log_2{(x+1)}=-1\]

And solve. But One doubt I have is whether you have take either union or intersection of the answers. @KingGeorge , @zepp , Any help in this part :)

Answer 67

@KingGeorge , Check again :D You are going wrong.

Answer 68

Ignore my ramblings.

Answer 69

x = 1 or -0.5?

Now I'm trying to understand why my method didn't work :|

Answer 70

Ok. The mistake you are making is that it should be

\[\large \log_a( \frac{b}{c}) = \log_a b - \log_a c\]

Answer 71

But in your case, you have taken it as
\[\large \frac{\log_a( b)}{\log_a (c)} \neq \log_a b - \log_a c\]

Answer 72

I'm good at this. I promise! Except when I do it around other people :(

Answer 73

LOL @KingGeorge . It happens many a time with me too. But for me, this happens with topic :integration :D

Answer 74

meh, I still don't see my mistake :|

Answer 75

OK. :D Let me have a check :)

Answer 76

Ah thanks :D

Answer 77

Your first step itself is a mistake.
It should be

\[\huge (x+1)^{\log_{2}(x+1)^8}=2^8\]

Answer 78

Oh, right :x

Answer 79

Herp derp, I probably thought that \(log_2 y = 8\) y = 3

Answer 80

No problem . Remember

\[\large \log_a b = c\]
Then
\[\large b = a^c\]

Answer 81

Haha, I know those properties, it's probably because of the 2 and the 8 :P
Anyway, thanks guys, it's quite surprising how much we can learn from 1 question! ;D

Answer 82

Yes. Thanks to @Directrix for the question and both of you @KingGeorge @zepp :)

Answer 83

Yeah, thanks to @Directrix and @KingGeorge :)

Answer 84

and @shivam_bhalla, of course!

Answer 85

Another Log Practice Problem For This Tutorial:
See Attachment.

Answer 86

Kellogg! :D

lol let me do it :)

Answer 87

I guess the first step would be \(\large \log_k 16 = \log_2 5\)
Then \(\huge \frac{\log_2 16}{log_2 k}=log_2 5\)
\(\huge \frac{4}{log_2 5} = log_2 k\)
\(\large k \approx 3.300549\)

Plug it in the question and the answer would be 625.