Logarithm help

[Dillinger4]

I know at at one point in time i am going to have to deal with logarithms and i think it's going to be semester after next so i wanted to see if i could get some help with them. Just the basic concepts.

A logarithm is the exponent of the power to which a base number must be raised to equal a given number.

x = a^y --> Log(a)^x = Log(a)^ay -->
Log(a)^x = y

8 = 2³ --> Log(2)⁸ = Log(2)²³ -->
Log(2)⁸ = 3

So when it is said the exponent of the power are we talking about the exponent of x = a^y?

[twanvl]

x = a^y <-> Log(a)^x = y
is not true (or perhaps this is a notation issue), the rule is:
x = a^y <-> _alog(x) = y
So:
8 = 2³ <-> ₂log(8) = 3

When x = a^y there are three things you can caluclate
given a,y find x: exponentiation
given x,y find a: root
given x,a find y: logarithm

[Dillinger4]

Yes yes much clearer now. So based on the definition of a logarithm
would it be correct to say that we are looking for the approximation of the power?

81 = 9² --> 9^log(81) = 2 --> log(81) = 1.908......... -->
9² = 81

[twanvl]

Not exactly, ₉log(81) means base-9 logarithm of 81, not 9^log(81) which means 9 raised to the base-e logarithm of 81.
When a base is not specified it is usually assumed to be equal to e (2.71828....), so log(81) is acutally _elog(81) = 1.908... this number is not related to ₉log(81) = 2.
The subscript notation for the base can be confusing sometimes, some people prefer to use superscript, as in ⁹log(81) = 2

[Dillinger4]

Ok say we have log(81). Since no based is specified we are using the natural log in the form of e^log(x) or ln x where e = 2.718......... So we have 2.718^log(81) = 1.908..... Now how is 1.908..... obtained?

[twanvl]

NOTE ^elog(81) is actually 4.3944... Your calculator's log function returns ¹⁰log(81)
A logarithm without base can mean many different things, to mathematicians the base is e, to computer scienctists the base is 2, to economists the base is 10...

To a mathetaician ¹⁰log(81)=1.908... is not obtained somehow, it is just the eternal truth.
To a computer or a person who wants to see the actual number it is more difficult. There is no easy way to find a logarithm. There are algorithms that can calculate them, but that is not something you want to do yourself. A simple algorithm you could use yourself is simply trying:

   10log(81) = x
   10^x = 81
try 1:
   10^1 = 10  (too small)
try 2:
   10^2 = 100 (too large)
try 1.5:
   10^1.5 = 31.. (too small)
try 1.75:
   10^1.75 = 56.2... (too small)
etc.

So basicly log(81)=1.908... because your calculator says so and because 10^log(81)=81

Your calculator probably does not have a logarithm key for any base, in that case you can use the formula:
^blog(x) = log(x)/log(b)
This works regardless of the base of the logarithm used.

[Dillinger4]

Posted by twanvl

So basicly log(81)=1.908... because your calculator says so and because ¹⁰log(81)=81

Is that correct? ¹⁰log(81)=81 -->
81 != 10⁸¹.

[Dillinger4]

I think you ment. ¹⁰log(81) == log(81) if i use my calculator.