math proofs

False, therefore [math]x^p\,=\,p\,\ln\,x[/math], QED.
False, therefore [math]\ln\frac xy\,=\,\ln\,x\,-\,\ln\,y[/math], QED.

>>9158921
ln(x) = a where e^a = x
(e^a)^p = x^p
e^ap = x^p
therefore
ln(x^p) = ap
ln(x^p)/p = a
ln(x^p)/p = ln(x)
ln(x^p) = p ln(x)

and

ln(x/y) = a where e^a = x/y
set e^p = x and e^q = y
(e^p)/(e^q) = e^(p-q) = e^a
therefore
ln(x/y) = p - q
ln((e^p)/(e^q)) = p - q
by definition ln(e^p) = p and ln(e^q) = q so if ln((e^p)/(e^q)) = p - q then ln(x/y) must = ln(x) -
ln(y)

>>9158921>>9158921
I never did a late transcendentals approach but I guess I can try, but I am a number theorist so I use log. Hope you dont mind. By definition:
[math] \log x = \int_1^x\frac{dt}{t} [/math].

The two properties you want to prove are actually just special cases of the property [math] \log xy = \log x + \log y [/math] so I'll just go for that.

Assuming [math] x \leq y [/math], [math] \log xy = \int_1^{xy}\frac{dt}{t} = \int_1^{x}\frac{dt}{t} + \int_x^{xy}\frac{dt}{t} = \log x + \int_x^{xy}\frac{dt}{t} [/math].

So now it is only necessary to find what [math] \int_x^{xy}\frac{dt}{t} [/math] is. Consider the substitution [math] u = \frac{t}{x} [/math]. Then the integral becomes [math] \int_1^{y}\frac{du}{u} = \log y[/math].

That proves the theorem. Cute. But late transcendentals is still gay.