Y tho

y tho

a^x * a^y = a^x+y

[math]log_{a}y=logay [/math]

lel, gay

Gay xD

something about x's and y's turn me on.
maybe thats why I like math so much, it's seXY

Properties of Logs
Def of Addition for Logs|

[math]For:\\ \quad \forall log_a x + log_a y \\Then:\\ \quad \rightarrow log_a xy[/math]

[math] \ln(x) + \ln(y) = \ln(x*y) [/math]
[math] u = \ln(x) [/math]
[math] v = \ln(y) [/math]
[math] e^{u} = e^{\ln(x)} [/math]
[math] e^{u} = x [/math]
[math] e^{v} = e^{\ln(y)} [/math]
[math] e^{v} = y [/math]
[math] e^{u}*e^{v} = x*y [/math]
[math] e^{u + v} = x*y [/math]
[math] \ln(e^{u + v}) = \ln(x*y) [/math]
[math] u + v = \ln(x*y) [/math]
[math] \ln(x) + \ln(y) = \ln(x*y) [/math]

this

lol gay

Awww, it's sharing :]

Just work with log10 and get back to basics with what the funny words and symbols are actually asking.

log10(10) + log10(100) = log10(10 * 100) = log10(1000) because:

A) log10(10) is asking how many times do you multiply 10 to get to 10?
B) log10(100) is asking how many times do you multiply 10 to get to 100?
C) log10(10 * 100) is asking how many times do you multiply 10 to get to 10 * 100?

So because you keep on asking how many times do you multiply the base (in this case 10) to get to the number you're plugging into that function known as the "logarithm," multiplying inside the parentheses of your function is equivalent to adding same base outputs of that function. Inside the parentheses multiplication is just multiplication, but outside when you're adding outputs of that function together you're on a different level where those outputs are answering questions about how many times to multiply the base by itself. So adding in that outside context is equivalent to multiplying in that inside context.

logay=loglmao

n=exp(m)=exp(log(n))
m=log(n)=log(exp(m))

log(x)+log(y)=log(exp(log(x)+log(y)))
=log(exp(log(x))exp(log(y))
=log(xy)

log10(10) = [math]\log_10 (10)[/math]
PLL

It's defined that way

Because n^a x n^b = n^(a+b)

what you say is partially untrue. n, a and b are relative numbers right ? here we discuss real numbers, which is different.

* define the function exp as the function f such as (i) f'=f (ii) f(0)=1 (cauchy theorem assures there is a unique solution to this problem)

* show then that exp(x+y)=exp(x) * exp(y)

* define then the function log as the inverse function of exp (inverse as : log o exp = id )

* deduce from this that log(xy) = log(x) * log(y)

* deduce immediately the equality you're asking help for with the very definition of log_a : log_a (x) = log(x) / log(a)

hth

Mistyped, I meant : log(xy) = log(x) + log(y)

An example to reverse a power:
[math]
c=a^b\\
a=c^{\frac{1}{b}}\\
b=\log _a\left(c\right)
[/math]
Explanation:
[math]
\log _{10}\left(a\right)+\log _{10}\left(b\right)=\log _{10}\left(a\cdot b\right)\\
10^a\cdot 10^b=10^{\left(a+b\right)}
[/math]
How do you do fellow kids?

>children
FTFY

looool it says gay, math is so cool xD

ermmm hi im not gay but i identify myself as a girl thuogh im actually a gyuy
so pzl dont ues the wrod gay cuz ist offends me :(
im only atracted 2 guys n i lkie thier bulges... ive become a tarp olny for sexual purposse ;_;
ddadi fedes me cumimes everydya cuz hes my mastuh n im a godo

log (base e) can also be defined as the integral of 1/x from 1 to t. So we find,

[math]\int_1^x \frac{1}{z} dz + \int_1^y \frac{1}{z} dz = \int_1^x \frac{1}{z} dz + \int_x^{xy} \frac{1}{xz} xdz = \int_1^{xy} \frac{1}{z} dz [/math]

shut up faggot

And, given a definition of ln, we set [math]\log_a(b) = \frac{\ln(b)}{\ln(a)}[/math], whereby [math]\log_a(x) + \log_a(y) = \frac{\ln(x)}{\ln(a)} + \frac{\ln(y)}{\ln(a)} = \frac{\ln(x) + \ln(y)}{\ln(a)} = \frac{\ln(xy)}{\ln(a)} = \log_a(xy)[/math].

>log (base e) can also be defined as the integral of 1/x from 1 to t.
Fair enough, this is the other classical equivalent definition. Question of tastes.

But then, instead of clumsily manipulating integrals, I prefer the following proof.

Let y>0. The function defined on ]0,+infty[ : x --> log(xy) is derivable and its derivative is y.1/(xy) = 1/x = log'(x). Two functions having same derivative differ only by a constant, hence: log(xy)=log(x)+k. Finding k is nothing difficult : previous equality taken in x=1 yields k=log(y).

>clumsily
cmon now, it's probably the easiest integral proof there is.

> ]0,+infty[
Your notation is shit, kys.

>notation
I can't do LaTeX here, multifag, but I do have smooth notations when my favorite packages perform the \infty command.

Plus, I'd rather have a shit notation than a shit reasoning.

So you, go KYS for reacting like a weak pussy and discussing stupid details instead of discussing maths.
(oh, btw, you nigger won't criticize notations from now on because:
>fag says integral of 1/x from 1 to t
>fag writes integral of 1/z from 1 to x
>fag doesn't know nobody uses 1/z dz except when doing holomorphic calculations
>fag changes the boundaries of integrals without any justification - is that a variable change? have you justified why it works?
)