Expert Answers to Algebra 2 Problems

Is there a proof that $n^x{m}^x=(n^x{)}^{(\log <!--\; \; -->(mn)/\log <!--\; \; -->(n))}$?
This isn't a homework question, just something I'm curious about, but you can treat it that way if you like.
So the other day I was playing with my calculator and I noticed that
$2^x{10}^x=(2^x{)}^{(\log <!--\; \; -->(20)/\log <!--\; \; -->(2))}$
I tried it out with some other numbers and came to the conclusion that
$n^x{m}^x=(n^x{)}^{(\log <!--\; \; -->(nm)/\log <!--\; \; -->(n))}$
So I wanted to see if there is a way to prove that.
I already know that $m=n^{(\log <!--\; \; -->(m)/\log <!--\; \; -->(n))}$ and I figured that there must be a relation. So from that I can see that $mn=n^{(\log <!--\; \; -->(nm)/\log <!--\; \; -->(n))}$. However I don't understand why that would mean that $(nm{)}^x=(n^x{)}^{\log <!--\; \; -->(nm)/\log <!--\; \; -->(n)}$.
Is what I say actually true? How do the powers fit into the proof?

logarithmic derivative of $x^{e^{(x^2+\cos <!--\; \; -->x)}}$
I'm having a hard time taking the derivative of
$f(x)=x^{e^{(x^2+cosx)}}.$
I'm aware that I have to take the logarithm of both sides.
$\ln <!--\; \; -->(y)=\ln <!--\; \; -->(x^{e^{(x^2+\cos <!--\; \; -->x)}})=\ln <!--\; \; -->(x)\cdot <!--\; \cdot \; -->e^{(x^2+\cos <!--\; \; -->x)}$
Which I tried to untie, so lets start: First I use the product rule:
$\frac{1}{y}\cdot <!--\; \cdot \; -->y^{\prime }=\frac{1}{x}\cdot <!--\; \cdot \; -->e^{(x^2+\cos <!--\; \; -->x)}+\ln <!--\; \; -->(x)\cdot <!--\; \cdot \; -->e^{(x^2+\cos <!--\; \; -->x)}$
Next the power rule:
$\frac{1}{y}\cdot <!--\; \cdot \; -->y^{\prime }=\frac{1}{x}\cdot <!--\; \cdot \; -->e^{(x^2+\cos <!--\; \; -->x)}+\ln <!--\; \; -->(x)\ast <!--\; \ast \; -->e^{(x^2+cosx)}\cdot <!--\; \cdot \; -->x^2+\cos <!--\; \; -->x\cdot <!--\; \cdot \; -->e^{(x^2+\cos <!--\; \; -->x)-<!--\; -\; -->1}$
Then I bring the y to the right:
$y^{\prime }=y\cdot <!--\; \cdot \; -->(\frac{1}{x}\cdot <!--\; \cdot \; -->e^{(x^2+\cos <!--\; \; -->x)}+\ln <!--\; \; -->(x)\cdot <!--\; \cdot \; -->e^{(x^2+\cos <!--\; \; -->x)}\cdot <!--\; \cdot \; -->x^2+\cos <!--\; \; -->x\cdot <!--\; \cdot \; -->e^{(x^2+\cos <!--\; \; -->x)-<!--\; -\; -->1}\cdot <!--\; \cdot \; -->2x-<!--\; -\; -->\sin <!--\; \; -->x)$
And exchange y with the term:
$y^{\prime }=x^{e^{(x^2+\cos <!--\; \; -->x)}}\cdot <!--\; \cdot \; -->(\frac{1}{x}\cdot <!--\; \cdot \; -->e^{(x^2+\cos <!--\; \; -->x)}+\ln <!--\; \; -->(x)\cdot <!--\; \cdot \; -->e^{(x^2+\cos <!--\; \; -->x)}\cdot <!--\; \cdot \; -->x^2+\cos <!--\; \; -->x\cdot <!--\; \cdot \; -->e^{(x^2+\cos <!--\; \; -->x)-<!--\; -\; -->1}\cdot <!--\; \cdot \; -->2x-<!--\; -\; -->\sin <!--\; \; -->x)$
This is extremely overwhelming for me and I have absolutely no clue if this is right, I looked in to the result of wolfram and It doesn't seem to be correct. Any help would be appreciated.

What are the basic, fundamental concepts of logarithms?
Soon I will be learning of logarithms, but I wish to have a head start over my class mates or at least a fair greeting to the concept. I would appreciate a very clear-cut answer with very straight to the point examples, an answer like this and with steps would be tremendously helpful and quickly voted best answer. Some specific questions on which to base your answer could be: why are they used? What is an example of a simple logarithm? how do you solve one?, and how does it apply to quadratic equations? Thanks in advance.

I recently came across the result that
$\underset{n=2}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{n^4-<!--\; -\; -->n^3+n+1}{n^6-<!--\; -\; -->1}=\frac{1}{2}$
I am wondering how one could proof this, generally how one could evaluate a sum over rational functions.
If I plug the sum into Wolfram Alpha it gives
$\frac{3k^4-<!--\; -\; -->k-<!--\; -\; -->2}{6k(k+1)(k^2+k+1)}$
as the $k$-th partial sum. Taking the limit as $n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$, this would in fact proof the upper equality.
Sadly, I could not wrap my head around how to get to Wolfram Alphas partial sum result. If anyone has an idea let me know. Any tips are appreciated.

Express $\lg <!--\; \; -->\frac{y}{x^2}$ in terms of a and b.
If $\lg <!--\; \; -->x^{2}y=a$ and $\lg <!--\; \; -->\frac{x}{y}=b$,Express $\lg <!--\; \; -->\frac{y}{x^2}$ in terms of a and b.
I did this, I don't know whether it right or wrong!
$\lg <!--\; \; -->x^{2}y=a$ can be written as: $lg{x}^2\cdot <!--\; \cdot \; -->\lg <!--\; \; -->y=a$
Then: $\lg <!--\; \; -->x^2=\frac{a}{\lg <!--\; \; -->y}$
and $\lg <!--\; \; -->\frac{x}{y}=b$ can be written as: $\lg <!--\; \; -->y=-<!--\; -\; -->b+\lg <!--\; \; -->x$
As I'll have to express $\lg <!--\; \; -->\frac{y}{x^2}$, I made $lg{x}^2$ and $\lg <!--\; \; -->y$, so now I can substitute the values into $lg\frac{y}{x^2}$
Is this the right way?
***If I continue in this way, the answer doesn't come right! :'( Help

There is a theorem that says rational functions in the extended complex plane are exactly the meromorphic functions.
After this, my textbook draws the corollary: "...as a consequence, a rational function is determined up to a multiplicative constant by prescribing the locations and multiplicities of its zeros and poles."
I don't see how it follows. Plus I think I have an example. Take a rational function with no zeros and two poles of multiplicity one, at 0 and 1.
At least two such functions exist:
$f(x)=\frac{1}{x}+\frac{1}{x-<!--\; -\; -->1}$
$g(x)=\frac{1}{x}+\frac{2}{x-<!--\; -\; -->1}$

Let U be a simply connected domain and let $f$ be a meromorphic function on U with only finitely many zeroes and poles. Prove that there is a holomorphic function g : U $\to <!--\; \to \; -->\mathbb{C}$ and a rational function $q$, such that
$f(z)=e^{g(z)}q(z)$

Suppose we have an integral (reduced and irreducible) affine curve $X$ over an algebraically closed field $k$. Suppose $t=\frac{f}{g}$ is a rational function on $X$ such that there exists an $n\in <!--\; \in \; -->\mathbb{N}$ such that $t^n=(\frac{f}{g}{)}^n$ is a regular function, i.e., $t^n\in <!--\; \in \; -->\mathrm{\Gamma <!--\; \Gamma \; -->}(X,{\mathcal{O}}_X)$ (the coordinate ring or ring of regular functions on $X$). Does this imply that $t$ itself is a regular function?
Note that if the curve is non-singular, this is immediate, as in this case $\mathrm{\Gamma <!--\; \Gamma \; -->}(X,{\mathcal{O}}_X)$ is a UFD and hence $b^n|a^n$ implies $b|a$ in $\mathrm{\Gamma <!--\; \Gamma \; -->}(X,{\mathcal{O}}_X)$. But is this true in general for an integral affine curve?

show that $(1+\frac{x}{n}{)}^n<e^x$ and $e^x<(1-<!--\; -\; -->\frac{x}{n}{)}^{-<!--\; -\; -->n}$ if $x<n$
If $n$ is a positive integer and if $x>0$,show that
$(1+\frac{x}{n}{)}^n<e^x$ and that $e^x<(1-<!--\; -\; -->\frac{x}{n}{)}^{-<!--\; -\; -->n}$ if $x<n$
I proved the first one by the inequalities
$(1+\frac{x}{n}{)}^n=\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{k})\frac{x^k}{n^k}\le <!--\; \le \; -->\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{x^k}{k!}<e^x$
the second one is equal to $(1-<!--\; -\; -->\frac{x}{n}{)}^n<e^{-<!--\; -\; -->x}$ but I feel I couldn't find a way to start.
In a previous exercise I proved
$(-<!--\; -\; -->1{)}^n{e}^{-<!--\; -\; -->x}<(-<!--\; -\; -->1{)}^n\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^k\frac{x^k}{k!}$
it seems to have some relation with the second question.So I hope someone could give me a hint. Thanks in advance.

I've recently been introduced to calculus and ran into a rational function that seems to be somewhat complicated for my understanding, which is shown in the form:
$f(x)=\frac{(x-a)(x-b)}{(c-a)(c-b)}\cdot <!--\; \cdot \; -->a+\frac{(x-b)(x-c)}{(a-b)(a-c)}\cdot <!--\; \cdot \; -->b+\frac{(x-a)(x-c)}{(b-a)(b-c)}\cdot <!--\; \cdot \; -->c$
Bearing in mind that when f(a)=b, f(b)=c and f(c)=a due to substitution.
However, the part I don't understand is the simplification of the equation into the quadratic form as follows:
$\frac{1}{(a-<!--\; -\; -->b)(a-<!--\; -\; -->c)(b-<!--\; -\; -->c)}[(a-<!--\; -\; -->b)a+(b-<!--\; -\; -->c)b+(c-<!--\; -\; -->a)c]x^2+[(a-<!--\; -\; -->b)b^2+(b-<!--\; -\; -->c)c^2+(c-<!--\; -\; -->a)a^2]x+[(a-<!--\; -\; -->b)a^2+(b-<!--\; -\; -->c)b^{2}c+(c-<!--\; -\; -->a)a{c}^2]$
Based on my algebraic knowledge, I know that the factorization formula is $x^2+(a+b)x+ab$
My question is how did the coefficients and rearranged dominator of the original equation come about? Can anyone detail it for me, please?

Prove that ${\log }_2<!--\; \; -->3$ is irrational
Seemingly simple homework assignment. Was never the best with logarithms, how would I go about proving?

Lower bound for logarithm?
Given a real $c$ such that $1<c$, is there any known and direct lower bound, other than $0$, for $(\ln <!--\; \; -->c)$, i.e., $A<\ln <!--\; \; -->c$
Thanks

I know that for polynomials $P,Q$, the equation $P(z)\equiv <!--\; \equiv \; -->Q(z)$ implies that they are of the same degree and have the same coefficients. Is there an analogous result for rational fucntions? That is, if $R,S$ are two rational functions and $R(z)=S(z)$ for all $z$ what is the relationship between $R$ and $S$?

Asset Depreciation - Logarithms
How many years would it take for the value of a car purchased at 30 000 to fall to 15 000 if it depreciates by 15% in the first year and 12% every year after that? This is not actually graded homework, but I solved it iteratively (compounding down to 26000 at 15% and so on) and would like to solve it using logarithms to find n as well. The answer is 5 years based on the iterative method at which point it is 14800 or so.

What is the domain of this function? (Don't know how to solve it, logarithms...)
Please explain how you solved it, thanks.
$f(x)=\sqrt{{\log }_x<!--\; \; -->2-<!--\; -\; -->{\log }_2<!--\; \; -->x}$

Let $X$ be a locally Noetherian scheme and let $f$ be a rational function on $X$ (i.e. the equivalence class of a pair $(U,f)$, where $f\in <!--\; \in \; -->{\mathcal{O}}_X(U)$ and $U$ contains the associated points of $X$, under obvious equivalence relation).
While reading Vakil's notes I wondered how could we define poles of such a rational function. After some thought I came up with the following definition: I'd say that a regular codimension one point $p$ is a pole if it's not in the domain of definition of $f$. If $X$ is also an integral scheme (or at least if all the stalks of ${\mathcal{O}}_X$ are integral domains, in which case we can cover $X$ with integral schemes), then this definition would coincide with the usual one, namely using the discrete valuation at $p$.
But there is something unnatural about my definition, since I was not able to relate the rational function with the discrete valuation on ${\mathcal{O}}_{X,p}$ and consequently was not able to determine the order of the pole. So I'd like to know if it's possible to define a meaningful notion of poles for rational functions on locally Noetherian schemes and how would it relate to my definition. By extension, consider the same question about zeros.

Noncircular construction of $e$ and $\ln$ for the real line
Could anyone direct me to (or possibly detail) a construction of $e$ and $\ln$ along the reals?
For example, they can define $e=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(1+\frac{1}{n}{)}^n$ but from this definition how do they prove:
It converges!
$\frac{d}{dx}{e}^x=e^x$
etc.!
Then if we know $e^x$ injective from $\mathbb{R}\to <!--\; \to \; -->{\mathbb{R}}^{\mathbb{+}}$, we can call $\ln <!--\; \; -->(x)$ the inverse of it. If we can prove $\ln$ is differentiable on its domain, then we can say:
$1=\frac{d}{dx}x=\frac{d}{dx}{e}^{\ln <!--\; \; -->(x)}=e^{\ln <!--\; \; -->(x)}\cdot <!--\; \cdot \; -->\ln <!--\; \; -->(x{)}^{\prime }=x\cdot <!--\; \cdot \; -->\ln <!--\; \; -->(x)\Rightarrow <!--\; \Rightarrow \; -->\ln <!--\; \; -->(x{)}^{\prime }=x^{-<!--\; -\; -->1}$
but this all depends on the above.

Example
$\frac{6x+2}{x^2-<!--\; -\; -->9}=\frac{6x+2}{(x+3)(x-<!--\; -\; -->3)}$
I know how to find the vertical and horizontal asypmtotes and everything, I just don't know how to find end behavior for a RATIONAL function without plugging in a bunch of numbers.