Expert Answers to Algebra 2 Problems

Solving a logarithmic equation. I need help on solving them when they are in exponents.
$x^{2{\log }^3<!--\; \; -->x-<!--\; -\; -->3/2\log <!--\; \; -->x}=\sqrt{10}$
Can someone help me to solve it?
Also, when we have $2{\log }^2<!--\; \; -->x$, is it equal to $4(\log <!--\; \; -->x{)}^2$

Let $f(x)=p(x)/q(x)$ be a rational function. Let q have a root at some x=a. It is clear that if $q(x)$ divides $p(x)$ then the limit of $f$ as $x$ goes to a exists. But is the converse true as well?

Recall that an elliptic curve over a field $k$ i.e a proper smooth connected curve of genus 1 equipped with a distinguished $k$-rational point, I'll be really grateful for any help in understanding the following part of our course
Let $(E,0)$ be an elliptic curve, using Riemann-Roch we construct an isomorphism into $\mathrm{Proj}k[X,Y,Z]/Y^{2}Z+a_{1}XYZ+a_{3}Y{Z}^2-<!--\; -\; -->X^3-<!--\; -\; -->a_2{X}^{2}Z-<!--\; -\; -->a_{4}X{Z}^2-<!--\; -\; -->a_6{Z}^3$ that can be written informally as $P\to <!--\; \to \; -->[x(P):y(P):1(P)]$, where $x$ and $y$ are rational functions such that $v_0(x)=-<!--\; -\; -->2$ and $v_0(y)=-<!--\; -\; -->3$.
Why does 0 map to the infinity point $O=[0:1:0]$? According to Hartshorne it is because both $x$ and $y$ have poles in 0 but I can't see why.

Could any one give me a hint how to show Every rational function which is holomorphic on every point of Riemann Sphere( ${\mathbb{C}}_{\mathrm{\infty <!--\; \infty \; -->}}$) must be constant?(with out applying Maximum Modulas Theorem). Thanks.

I know how to find the oblique/slant asymptote of a rational function. What I'm unsure about, because I've never dealt with this, is how many oblique asymptotes can a function actually have? And is it only possible to have an asymptote if the degree of the numerator polynomial and the degree of the denominator polynomial only differ by 1?

Inequality with logarithms
How do I show that
$\frac{1}{n-<!--\; -\; -->1}\ge <!--\; \ge \; -->\ln <!--\; \; -->\left(\frac{n}{n-<!--\; -\; -->1}\right)$
for n>1?
As far as I can tell, exponentiating both sides with base e won't help, because then I get a nasty term on the LHS.

Make derivations over sums
I have this kind of sums
${(\underset{i_1=0}{\overset{4}{\sum <!--\; \sum \; -->}}\underset{i_2=0}{\overset{4}{\sum <!--\; \sum \; -->}}\log <!--\; \; -->(f(X,i_1,i_2)))}^{\prime }$
And we want to derive in respect to
$x_i$
, which is an element of the vector X.
How I should do that? I did basic maths at Highschool, but I do not know how to derive in respect to an element which belongs to a vector, at a function that has sums as well.
Could you help me with that?

How do computers calculate the log of a value?
I'm not sure if this question belongs on StackOverflow or here (please let me know if the former, and i'll delete this and ask there), but I was wondering how the log or ln of a value is calculated computationally with accuracy? Is some series implemented that approximates a value?
I looked into the Taylor series for the natural logarithm, but that is apparently only accurate for 0 < x < 2, and I can't find anything else. I tried looking for source code for Java's Math#log function as well to see what algorithm they implemented, but couldn't find any since it's implemented in a native language rather than in Java.

I sometimes have some trouble finding the range of a rational function. What is an easy or good way to do this?

Closed form for ${\int <!--\; \int \; -->}_0^1\sqrt{\frac{2-<!--\; -\; -->x}{(1-<!--\; -\; -->x)x}}\log <!--\; \; -->\left(\frac{(2-<!--\; -\; -->x)x}{1-<!--\; -\; -->x}\right)dx$
Is it possible to find a closed form for this integral?
$Q={\int <!--\; \int \; -->}_0^1\sqrt{\frac{2-<!--\; -\; -->x}{(1-<!--\; -\; -->x)x}}\log <!--\; \; -->\left(\frac{(2-<!--\; -\; -->x)x}{1-<!--\; -\; -->x}\right)dx$

How can I create a rational function g(x) with these requirements:
1. The vertical asymptotes are at 2 and 5
2. The function has a slant asymptote y=2x+1
3. The y-intercept is 7

How to write in $2^x=5$ logarithmic form?

Properties of Natural Logarithm I need help finding the Derivative
$y=\ln <!--\; \; -->(x{)}^2$
I am not sure why the answer would be $\frac{2\ln <!--\; \; -->(x)}{x}$
I used this property "power rule" "$\ln <!--\; \; -->(x^n)=n\ln <!--\; \; -->(x)$
So i got $2\ln <!--\; \; -->(x)$
the derivative of that using the constant multiplier rule i got
$\frac{2}{x}$
can I use the other chain rule to $y=f(u)$ and $g=g(x)$ Am i not supposed to bring that 2 in front becuase the whole expression is getting raised not the x? Any help would be great,

The given function is
$f(x,y)=\frac{1}{x^2+y^2-<!--\; -\; -->1}$
From my understanding, a rational function such as this one would not have any points in which $\mathrm{\nabla <!--\; \nabla \; -->}f(x,y)=\mathbf{0}$, so I would conclude that no such points exist. Would this be true or am I not looking at this in a different way?

How do I prove that the limit as one function goes to infinity is equal to another function?
I was playing around with the integral $\int <!--\; \int \; -->x^{n}dx$ and noticed that it is always $\frac{x^{n-<!--\; -\; -->1}}{n-<!--\; -\; -->1}+c$ except for the singular case $n=-<!--\; -\; -->1$. So I could pick n arbitrarily close to $-<!--\; -\; -->1$ and the formula works, but as soon as I hit $-<!--\; -\; -->1$ it breaks (so to speak)!
Here's my work thus far:
$\int <!--\; \int \; -->x^{-<!--\; -\; -->(n+1)/n}dx=-<!--\; -\; -->n{x}^{-<!--\; -\; -->1/n}+c$
Working with $n=100$ with the initial condition $y(1)=0$ I got:
$-<!--\; -\; -->100x^{-<!--\; -\; -->1/100}+c=0$
$-<!--\; -\; -->100+c=0$
$c=100$
So I figure the following could be true:
$-<!--\; -\; -->n{x}^{-<!--\; -\; -->1/n}+n\approx <!--\; \approx \; -->\ln <!--\; \; -->x$
How can I prove that as I take $n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$ the two functions are equivalent?
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}-<!--\; -\; -->n{x}^{-<!--\; -\; -->1/n}+n=\ln <!--\; \; -->x$
Or equivalently,
$e^{\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}-<!--\; -\; -->n{x}^{-<!--\; -\; -->1/n}+n}=x$

Solving inequality having log
I am struggling to solve this inequality involving logarithm. How to find out values of n for which below inequality holds good:
$\frac{{\log }_2<!--\; \; -->n}{n}>\frac{1}{8}$

I am trying to prove that
${\int <!--\; \int \; -->}_C\frac{P(z)}{Q(z)}dz=0$
If polynomial order of $Q$ is 2 or more than that of $P$, using the theorem stating that if a function has a finite number of singular points all interior to a contour $C$, then
${\int <!--\; \int \; -->}_{C}f(z)dz=2\mathrm{\pi <!--\; \pi \; -->}i\text{Res}[\frac{1}{z^2}f(\frac{1}{z})]$
I received the hint that, under the conditions described above, the rational function in the integrand can be written as a McLauran Series with no negative powers of z. This would imply that the residue is zero...
My problem is, I can't seem to wrap my head around how the rational function given to me can be written into the form of a power series with only positive exponents...
So, as the title says, When can a rational function be represented as a power series? I'm not looking for a full proof, but a few details would be nice, just so I feel more comfortable running with the hint.

Simplifying an expression using a logarithm
I have the following expression
$\frac{1}{1+\mathrm{\rho <!--\; \rho \; -->}}(1+n{)}^{(1-<!--\; -\; -->\mathrm{\sigma <!--\; \sigma \; -->})}\ast <!--\; \ast \; -->(1+{\mathrm{\gamma <!--\; \gamma \; -->}}_A{)}^{1-<!--\; -\; -->\mathrm{\sigma <!--\; \sigma \; -->}}<1$
and have to use logarithms to get the following
$(1-<!--\; -\; -->\mathrm{\sigma <!--\; \sigma \; -->})(n+{\mathrm{\gamma <!--\; \gamma \; -->}}_A)<\mathrm{\rho <!--\; \rho \; -->}$
Could someone please offer me some help on how to do it? That's a homework question in Macroeconomics - we got the solutions and were told to use the logarithms to get them. However, I'm struggling with the complete procedure, so any help is more than welcome.
Thanks!

Let's say we have a rational function $f$ (i.e polynom divided by polynom) , and assume that $f$ has no poles in the upper plane $\{z;Imz\ge <!--\; \ge \; -->0\}$
we have to prove that:
$sup\{|f(z)|;Imz\ge <!--\; \ge \; -->0\}=sup\{|f(z)|;Imz=0\}$
1. I assume that we are dealing in cases where the suprimum exists.
2. every approach I tried, I can't really understand what is the significant of function being rational. I know, for example, that there is finite number of zero's, and finite number of poles, but I can't see how it is helpful.

Algebra involving powers of two when solving for cardinality involving power sets
What's the proper way of solving this problem (self study)?
Find $m$ if $|A|=m$, $|B|=m+2$ and $|\mathcal{P}(B)|-<!--\; -\; -->|\mathcal{P}(A)|=48$
Since the numbers are small, I can get m=4 pretty much by inspection
I tried the more general approach of making the equation and solving for m, but got stuck because my algebra skills are weak:
$2^{(m+2)}-<!--\; -\; -->2^m=48$
However, ${\mathrm{l}\mathrm{o}\mathrm{g}}_2(48)$ is hanging me up. Plugging in $m=4$ does satisfy the equation, so I'm just messing up something really obvious here.