A logarithm is an arithmetic operation in a three-way system: exponents, radicals, and logarithms. There are two ways to write a logarithm, which are logₐc and ᵃlogc. They mean the same thing, just with different notations.
If we take a as a base, b as its exponent, and c as its result; then c is defined as aᵇ, a as ᵇ√c, and b as logₐc. This is the definition of a logarithm!
There are, arguably, two basic identities of logarithms.
The first one is logₐ1 = 0. This comes from the fact that any real number to the power of 0 will always equal 1.
Secondly, logₐa = 1. This comes from the fact that any number to the power of 1 will always be equal to itself!
You may encounter two logarithms that aren't written with the typical notation: one with e or 10 as its base. These numbers don't have any special property—they're just conventionally written differently. A logarithm with base e is written as "ln" and base 10 is just "log".
There are a few stuff you can do to manipulate logarithms—I'll discuss three rules and seven properties of logarithms in this post. Keep the equations in the second paragraph in mind!
Rule 1: the base of a logarithm must be a positive real number other than 1.
logₐc = b, 0<a<1 or a>1
Logarithms with negative bases can get pretty ugly quickly. It can only be calculated with specific numbers, or else it'll return imaginary numbers. Since logarithms are used at the real number line, negative bases don't get used.
0, on the other hand, is a unique case. No matter what real number you exponentiate it to, it'll always return 0(?). So is the case with 1.
Rule 2: the value inside a logarithm must be greater than 0.
logₐc = b, c > 0
With our base being a positive number, no matter what real number our power is, we'll never get a value equal to or less than 0.
Rule 3: the values inside of logarithms are equal if they have the same base.
if logₐb = logₐc, then b = c
Time for logarithm properties!
Firstly, a^(logₐc) = c. How do we prove this? As we've seen at the beginning of this post, logₐc is just b. So, we can substitute that into the equation.
a^b = c. Using the three equations in the second paragraph, we've proven that this property is true by definition! Next, logₐmn = logₐm + logₐn "Chat, is this real?" you may be asking yourself, or perhaps, "Source?". By exponentiating both sides with a as their bases, we get
mn = a^(logₐm + logₐn).
Using properties of exponents, we can split the right-hand term.
mn = a^(logₐm) ⋅ a^(logₐn) Using the property from the previous paragraph, we can simplify the right-hand side of the equation into m⋅n. mn = m⋅n Woah, there you go: the L.H.S. is equal to the R.H.S.!
Thirdly, logₐ(m/n) = logₐm - logₐn
This is proven similarly to the previous property, so I'll leave it as an exercise for the reader. ;)
Another one, logₐ(mⁿ) = n⋅logₐm
If you think about it, mⁿ is just a repeated multiplication of m, n times!
logₐ(mⁿ) = logₐ(m⋅m⋅m⋅...⋅m) Using the second property, we can split these m's into additions of logₐm's.
logₐ(mⁿ) = logₐm + logₐm + ... + logₐm Do you see it? Repeated addition is just multiplication! Hence, the addition of logₐm n times is just logₐm times n!
After that, logₐb = logₓb / logₓa Let's start with something simple.
b = b
We can turn the left-hand side b into a^(logₐb) using the first property.
a^(logₐb) = b
Then, we take the logarithm of both sides, with a new variable x as a base.
logₓa^(logₐb) = logₓb
Using the previous property, we can take the exponent out and turn it into a coefficient!
logₐb ⋅ logₓa = logₓb
With some basic algebra, we can turn this into the expression we started with! This has been, indeed, fact-checked by real math patriots.
Finally, logᵤ‸ᵥ(mⁿ) = n/v⋅logᵤm
To prove this property, we need to use the previous property. We can put our base and value (inside the L.H.S. logarithm) into two logarithms with base a, a sort of... "dummy" variable.
logᵤ‸ᵥ(mⁿ) = logₐ(mⁿ) / logₐ(uᵛ) We can then take the n and v out so they become coefficients.
logᵤ‸ᵥ(mⁿ) = (n⋅logₐm) / (v⋅logₐu) logᵤ‸ᵥ(mⁿ) = (n/v)⋅(logₐm / logₐu) Using the previous property once again, we can simplify (logₐm / logₐu) into logᵤm!
logᵤ‸ᵥ(mⁿ) = (n/v)⋅(logᵤm)
Hence, the property is proven!
That's all I have for today! I think this might be my longest post in Peridot haha. Oh yeah, Ken requested this post. Anyway, I hope you guys enjoyed reading it!