Author Topic: Is this working right/Random questions (Read 3653 times) Tweet Share

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 =/)

Mao · « **Reply #1 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.

cara.mel · « **Reply #2 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?

Mao · « **Reply #3 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...

cara.mel · « **Reply #4 on:** September 14, 2008, 09:56:28 am »

why are they like that

why does e^ln gfkjfdsgkj = gfkjfdsgkj

shinny · « **Reply #5 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.

cara.mel · « **Reply #6 on:** September 14, 2008, 10:30:37 am »

why can you log both sides and drop both logs

thank you =)

Flaming_Arrow · « **Reply #7 on:** September 14, 2008, 10:40:01 am »

$\log_e a = \log_e b$

so they both have same log base

$\implies a = b$

shinny · « **Reply #8 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

Mao · « **Reply #9 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 $a^k =b \implies \log_a b = k$

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

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

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

cara.mel · « **Reply #10 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

Captain · « **Reply #11 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.

$x log_a b = log_a c$

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

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

cara.mel · « **Reply #12 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)

Flaming_Arrow · « **Reply #13 on:** September 14, 2008, 11:07:44 am »

Mao · « **Reply #14 on:** September 14, 2008, 11:08:31 am »

yes

log is defined as the inverse of the logarithm function

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