ATAR Notes: Forum

Uni Stuff => Science => Faculties => Mathematics => Topic started by: cara.mel on August 21, 2008, 10:19:19 am

Title: Is this working right/Random questions
Post by: cara.mel on August 21, 2008, 10:19:19 am: I have lots of trouble with Log (not so much with Exponential, but that is hard too), I can visualise them fine but I don't understand any of the rules for it.

I am sorry if the latex is shite (it looks all squished up, ie the lines are particularly friendly towards each other)

$- ln(1 + e^{-y}) = x + \frac{x^2}{2} + c, where \; y(0) = 0$
$ln(1 + e^{-y}) = -x - \frac{x^2}{2} - c$
$1 + e^{-y} = e^{-x - \frac{x^2}{2} - c}$
$e^{-y} = e^{-x - \frac{x^2}{2} - c} - 1$

Making the substitution now because my friend said so:
$e^{0} = e^{-0 - \frac{0^2}{2} - c} - 1$
$1 = e^{-c} - 1$
$2 = e^{-c}$
$-c = ln(2)$ **is this right?

∴ (does Latex have therefore?)
$e^{-y} = e^{-x - \frac{x^2}{2} + ln(2)} - 1$
$-y = ln(e^{-x - \frac{x^2}{2} + ln(2)} - 1)$
$y = - ln(e^{-x - \frac{x^2}{2} + ln(2)} - 1)$

Then, as $e^{a + b} = e^a \cdot e^b$
$e^{\left(-x - \frac{x^2}{2}\right) + ln(2)} = e^{-x - \frac{x^2}{2}}\cdot e^{ln(2)} = 2e^{-x - \frac{x^2}{2}}$ *why does e^ln x = x
∴ $y = - ln(2e^{-x - \frac{x^2}{2}} - 1)$
Can this be simplified still?

Why does $e^{iwx} = cos (wx) + i \cdot sin (wx)$
Why does $\frac{e^x + e^{-x}}{2} = cosh (x)$ (I don't care if that last expression is actually sinh x, I am stupid and don't know which one has + and which one has - in it, I think I got it right =/)
Title: Re: Is this working right
Post by: Mao on August 21, 2008, 07:52:17 pm: for "squished up", you usually put extra line spacing when using fractions, etc

$c=-\log_e(2)$ , that is correct.

that expression for y does not appear like it can be simplified any further.

$e^{iwx}=\cos(wx)+i\cdot \sin(wx)$ is euler's formula. the proof for it can be found here: http://en.wikipedia.org/wiki/Euler's_formula#Using_calculus

the definition of the hyperbolic functions are the odd and even parts of the natural exponent, that's just the way it is defined. and you are right, that is the expression for cosh.
Title: Re: Is this working right
Post by: cara.mel on August 22, 2008, 09:41:09 am: Thank you mao

How can I learn rules for Exponential and Log.

Why are some things 'impossible' to solve algebraically?
Title: Re: Is this working right/Random questions
Post by: Mao on September 14, 2008, 09:36:11 am: $\log a + \log b = \log (ab)$

$\log a - \log b = \log \left(\frac{a}{b}\right)$

$\log 1 = 0$

$\log_a a =1$

$-\log b = 0 - \log b = \log \frac{1}{b}$

$\log \left(b^c\right) = c\cdot \log b$

$\frac{\log_a b}{\log_a c} = \log_c b$

thats all i think...
Title: Re: Is this working right/Random questions
Post by: cara.mel on September 14, 2008, 09:56:28 am: why are they like that

why does e^ln gfkjfdsgkj = gfkjfdsgkj
Title: Re: Is this working right/Random questions
Post by: shinny on September 14, 2008, 10:22:14 am: I'm not sure if theres a faster proof for that rule but here's one:
Let: $e^{log_e a}=b$

Log both sides: $\log_e e^{log_e a}=\log_e b}$

Chuck the power down using that other log law: $\log_e a \log_e e=\log_e b}$

Simplify the log: $\log_e a=\log_e b}$

Equate the logs: $a=b$

So therefore: $e^{log_e a}=a$

Oh, and if you're seriously struggling, try and derive all the log laws yourself through exponentials and it makes things alot more obvious why they're like that. My teacher for both years so far is really into proper understanding, so I've had to derive basically every formula I've used in spesh and methods so far, and yeh, it actually does help.

EDIT: Ok whoops, you mentioned you weren't too great at exponentials either. In that case, go back to the basic level and write them out as they are in basic form. i.e. 2^3=2*2*2, and especially being able to distinguish between (2^3)*(2^4) and (2^3)^4. Then experiment a bit more I guess and you'll see why things are like that =P Either that or just rote learn them and hope for the best. I find the exponential laws to be quite logical, but yeh, the log ones are kinda weird.
Title: Re: Is this working right/Random questions
Post by: cara.mel on September 14, 2008, 10:30:37 am: why can you log both sides and drop both logs

thank you =)
Title: Re: Is this working right/Random questions
Post by: Flaming_Arrow on September 14, 2008, 10:40:01 am: $\log_e a = \log_e b$

so they both have same log base

$\implies a = b$
Title: Re: Is this working right/Random questions
Post by: shinny on September 14, 2008, 10:40:54 am: Well for log'ing both sides, its just like with any function (such as square root, squaring etc); if you do it to both sides as a whole, then yeh, its still equal (some exceptions though I imagine). As for dropping powers...hmmm..I'll get a proof for that one, wait a sec =P

EDIT: ok mao beat me to it =T
Title: Re: Is this working right/Random questions
Post by: Mao on September 14, 2008, 10:44:07 am: the definition of a logarithm is the index the base must be raised to to get to that number,

i.e. if $a^b = c\implies \log_a c = b$ , the base "a" must be raised to the b^th power to get to c,

using this principle, if we raise the base by the power of the base:

$a^{\log_a c}=a^b = c$

and, $\log_a a^b = b$

also, from the few index laws:

$a^m \cdot a^n = a^{m+n}=c\implies \log_a c = m+n = \log_a a^m + \log_a a^n$ (this is the first log law)

the second one can be shown in a similar fashion.

the third law lies on the fact that $a^0=1$ for non-zero a. hence, the log of 1 of any non-zero base is 1.

the fourth law is evidently true: $log_a a=log_a a^1 = 1$

the fifth is a combinations of the previous

the next one uses $\left(a^m\right)^n = a^(m\cdot n)\implies \log_a (a^m)^n = n\cdot m = n\cdot \log_a a^m$

and the last one is tricky:
$c=a^m = b^n = a^{kn}$ , where k is a constant such that $a^k =b \implies \log_a b = k$

$\log_a c = \log_a a^m = \log_a a^{kn}$
$\implies m = kn$

$\implies \frac{m}{k}=n=\log_b b^n$

$\implies log_b c = \frac{\log_a c}{\log_a b}$
Title: Re: Is this working right/Random questions
Post by: cara.mel on September 14, 2008, 10:59:51 am: is there like a list which puts Exponential Rule next to corresponding Log rule?

LIke I remember e^a*e^b = e^(a+b) is related to log a + log b = log ab
what are the other ones paired up with
I don't think I know all the exponential rule but I know that one because if it was like e^2*e^3 = (e*e)*(e*e*e) = e^5
Title: Re: Is this working right/Random questions
Post by: Captain on September 14, 2008, 11:02:43 am: Quote
and the last one is tricky:
$c=a^m = b^n = a^{kn}$ , where k is a constant such that $a^k =b \implies \log_a b = k$

$\log_a c = \log_a a^m = \log_a a^{kn}$
$\implies m = kn$

$\implies \frac{m}{k}=n=\log_b b^n$

$\implies log_b c = \frac{\log_a c}{\log_a b}$

$log_b c = x$

We know that $b^x = c$

If we do the same thing to both sides of an equation, it stays the same.

$log_a$ both sides

$log_a b^x = log_a c$

Using log laws, which Mao has already proved.

$x log_a b = log_a c$

$\implies x = \frac{log_a c}{log_a b}$

$\implies log_b c = x = \frac{log_a c}{log_a b}$
Title: Re: Is this working right/Random questions
Post by: cara.mel on September 14, 2008, 11:06:37 am: thank you everyone :):)
so is Log like ^ and arcsin and stuff?

Eg
sin theta = 0.5
arcsin (sin theta) = arcsin (0.5)
theta = arcsin (0.5)
Title: Re: Is this working right/Random questions
Post by: Flaming_Arrow on September 14, 2008, 11:07:44 am: $\log_b x = y$

$x = b^y$
Title: Re: Is this working right/Random questions
Post by: Mao on September 14, 2008, 11:08:31 am: yes

log is defined as the inverse of the logarithm function
Title: Re: Is this working right/Random questions
Post by: cara.mel on September 14, 2008, 11:08:42 am: Quote from: chath on September 14, 2008, 11:07:44 am
$\log_b x = y$

$x = b^y$

I understand that much I dont know anything else :(
Title: Re: Is this working right/Random questions
Post by: cara.mel on September 14, 2008, 11:11:51 am: Quote from: Mao on September 14, 2008, 11:08:31 am
yes

log is defined as the inverse of the logarithm function

Does that mean arcsin is inverse of sin?
I understand trig except for how it is meant to be linked with exponential I don't follow that

edit: so if you pretend trig and exponential are seperate I think i will get trig example if you compare it to exponential example
Title: Re: Is this working right/Random questions
Post by: Flaming_Arrow on September 14, 2008, 11:18:53 am: Quote from: caramel on September 14, 2008, 09:26:21 am
Quote from: Mao on September 13, 2008, 11:58:51 pm
Also, a note (a tiny trick if you like), always use the log laws to avoid $\log_a \frac{1}{b}$ , they are very messy to deal with, it is far easier to use its equivalence $-\log_a b$
and, you should seek to express everything with one log term (having two or more is often confusing and difficult to deal with

How did you do that?
What are Log Law?
you can post reply here: http://vcenotes.com/forum/index.php/topic,4899.0.html

$\log_{10} 500 - \log_{10} 5 = x$

$\log_{10} (\frac {500}{1} * \frac {1}{5}) = x$

$\log_{10} \frac {500}{5} = x$

$100 = 10^{x}$

$10^2 = 10^x$

$\implies \boxed {x = 2}$
Title: Re: Is this working right/Random questions
Post by: /0 on September 14, 2008, 02:43:01 pm: Quote from: caramel on September 14, 2008, 11:06:37 am
thank you everyone :):)
so is Log like ^ and arcsin and stuff?

Eg
sin theta = 0.5
arcsin (sin theta) = arcsin (0.5)
theta = arcsin (0.5)

Yeah they're both inverses, in that respect they're analogous, but I think the properties of arcsin are more complicated (perhaps just because I haven't learnt them :P)

$e^{\theta} = 0.5$

$\log_e \left({e^{\theta}}\right) = \log_e \left(0.5\right)$

$\theta = \log_e 0.5$
Title: Re: Is this working right/Random questions
Post by: Mao on September 15, 2008, 04:19:36 pm: Quote from: chath on September 14, 2008, 11:18:53 am
Quote from: caramel on September 14, 2008, 09:26:21 am
Quote from: Mao on September 13, 2008, 11:58:51 pm
Also, a note (a tiny trick if you like), always use the log laws to avoid $\log_a \frac{1}{b}$ , they are very messy to deal with, it is far easier to use its equivalence $-\log_a b$
and, you should seek to express everything with one log term (having two or more is often confusing and difficult to deal with

How did you do that?
What are Log Law?
you can post reply here: http://vcenotes.com/forum/index.php/topic,4899.0.html

$\log_{10} 500 - \log_{10} 5 = x$

$\log_{10} (\frac {500}{1} * \frac {1}{5}) = x$

$\log_{10} \frac {500}{5} = x$

$100 = 10^{x}$

$10^2 = 10^x$

$\implies \boxed {x = 2}$

... non sequitur?
Title: Re: Is this working right/Random questions
Post by: dcc on September 24, 2008, 07:46:31 pm: Quote from: Mao on September 15, 2008, 04:19:36 pm
Quote from: chath on September 14, 2008, 11:18:53 am
Quote from: caramel on September 14, 2008, 09:26:21 am
Quote from: Mao on September 13, 2008, 11:58:51 pm
Also, a note (a tiny trick if you like), always use the log laws to avoid $\log_a \frac{1}{b}$ , they are very messy to deal with, it is far easier to use its equivalence $-\log_a b$
and, you should seek to express everything with one log term (having two or more is often confusing and difficult to deal with

How did you do that?
What are Log Law?
you can post reply here: http://vcenotes.com/forum/index.php/topic,4899.0.html

$\log_{10} 500 - \log_{10} 5 = x$

$\log_{10} (\frac {500}{1} * \frac {1}{5}) = x$

$\log_{10} \frac {500}{5} = x$

$100 = 10^{x}$

$10^2 = 10^x$

$\implies \boxed {x = 2}$

... non sequitur?

For positive numbers there is no worry.
Title: Re: Is this working right/Random questions
Post by: Mao on September 24, 2008, 09:25:58 pm: nono, as in

"What are the log laws?"

" $\log_{10}500 - \log_{10}5=x$ "
Title: Re: Is this working right/Random questions
Post by: /0 on September 24, 2008, 11:36:37 pm: Quote from: Mao on September 24, 2008, 09:25:58 pm
nono, as in

"What are the log laws?"

" $\log_{10}500 - \log_{10}5=x$ "

For an appropriate $x\in \mathbb{R}:x=2$