Expert Answers to Algebra 2 Problems

Find the ${\log }_y<!--\; \; -->(x)$given x and y in the form of a number.

So I was doing logs trying to reteach myself pre-calculus and I noticed I don't remember how to turn logs in to numbers. I tried to calculate the ${\log }_4<!--\; \; -->8$ so I turned it in to $4^x=8$ and then realized that $8=4^1\sqrt{4}=4^{1.5}$. Isn't there a method to calculate the answer even when it is an irrational number?

Find the partial fraction decomposition of the rational function ${\displaystyle \frac{2x^3+7x+5}{(x^2+x+2)(x^2+1)}}$
I have tried dividing first but keep running into problem after problem.

Find the domain of $g(x)=\ln <!--\; \; -->\left(\frac{x}{x-<!--\; -\; -->1}\right)$
I know the argument of the function has to be greater than 0, so:
$\begin{array}{rl}& \left(\frac{x}{x-<!--\; -\; -->1}\right)>0\\ & x>0\end{array}$
however in this case $x\ne <!--\; \ne \; -->1$, $x\ne <!--\; \ne \; -->0$ as they result in an answer which is undefined, so I think it's reasonable for me to say the domain is $x>1$, however this is not the correct answer.

Solving logarithms with different bases?
How would I go about getting an exact value for a question like: ${\log }_8<!--\; \; -->4$
I know that $8^{2/3}=4$ but how would I get that from the question?

Choosing a branch for $\log$ when comparing $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\prod <!--\; \prod \; -->}}(1+a_n)$ and $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\log <!--\; \; -->(1+a_n)$
On Ahlfors on p. 191 he is talking about the relation between $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\prod <!--\; \prod \; -->}}(1+a_n)$ and $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\log <!--\; \; -->(1+a_n)$. He says:
Since the $a_n$ are complex, we must agree on a definite branch of the logarithms, and we decide to choose the principal branch in each term.
He makes no mention of the possibility that one of the terms might lie on the negative real axis. What then?

Could someone please let me know if my steps are correct? I am trying to rewrite the first line as a rational function, $\frac{p(x)}{q(x)}$
$\begin{array}{rl}& 3x^1+3x^2+9x^3+9x^4+27x^5+27x^6+\cdots <!--\; \cdots \; -->\\ =& 3x(1+x)+9x^3(1+x)+27x^5(1+x)+\cdots <!--\; \cdots \; -->\\ =& 3x(1+x)(1+3x^2+9x^4+\cdots <!--\; \cdots \; -->)\\ =& (3x+3x^2)\frac{1}{1-<!--\; -\; -->3x^2}\\ =& \frac{3x+3x^2}{1-<!--\; -\; -->3x^2}\end{array}$

How do I solve, $3^{2x+1}-<!--\; -\; -->10\cdot <!--\; \cdot \; -->3^x+3=0?$
I am baffled to solve this equation. With graphing I have found the answers to be x=1 and x=-1. I would like to know how to solve this equation though.
I have tried many different approaches including rearranging to these various forms (in no particular order):
$3^{2x+1}-<!--\; -\; -->3^{x+2}-<!--\; -\; -->3^x+3=0$
$3^{2x+1}\cdot <!--\; \cdot \; -->(1-<!--\; -\; -->10\cdot <!--\; \cdot \; -->3^{-<!--\; -\; -->x-<!--\; -\; -->1}+3^{-<!--\; -\; -->2x})=0$
$-<!--\; -\; -->2x=10\cdot <!--\; \cdot \; -->3^{-<!--\; -\; -->x-<!--\; -\; -->1}-<!--\; -\; -->3^0$This equation has a term eliminated in it already.
The last equation written is the closest I have gotten to finding the answer, but I don't know how to proceed any further.

Let $C$ be the algebraic curve defined by the modular polynomial ${\mathrm{\varphi <!--\; \varphi \; -->}}_N$ of order $N>1$ over the rational numbers, i.e.
$C:=\text{specm}(\mathbb{Q}[X,Y]/{\mathrm{\varphi <!--\; \varphi \; -->}}_N(X,Y)).$
The singularities of this curve can be removed and we obtain a nonsingular curve $C^{sn}$n, then, we can embed $C^{sn}$ into a complete non-singular curve $\stackrel{¯<!--\; ¯\; -->}{C}$.
In Milne's notes "Modular Functions and Modular Forms" it is written:
The coordinate functions $x$ and $y$ are rational functions on $\stackrel{¯<!--\; ¯\; -->}{C}$, they generate the field of rational functions on $\stackrel{¯<!--\; ¯\; -->}{C}$, and they satisfy the relation ${\mathrm{\varphi <!--\; \varphi \; -->}}_N(x,y)=0$.
I assume, by coordinate functions he means the functions $f(X),g(X)\in <!--\; \in \; -->\mathbb{Q}[X]$ such that ${\mathrm{\varphi <!--\; \varphi \; -->}}_N(f(x),g(x))=0$ for all $x\in <!--\; \in \; -->Q$. However, I don't understand why the field of rational functions on $\stackrel{¯<!--\; ¯\; -->}{C}$ is generated by these functions. Could someone explain this to me?
Thank you very much in advance!

How can you prove this equality?
I am trying to figure out the following equality, but cannot seem to get anywhere. I tried integrating by parts, but that blew up when I set u = (log x)^n and tried to take log (0). I also tried differentiating the right side but got stuck when I did not know the derivative of $n!$
How can we prove that the following is true:
${\int <!--\; \int \; -->}_0^1{x}^a(\log <!--\; \; -->x{)}^{n}dx=\frac{(-<!--\; -\; -->1{)}^n(n!)}{(a+1{)}^{n+1}}$
Any suggestions would be greatly appreciated.

Evaluate ${\log }_5<!--\; \; -->{19}^2/({\log }_3<!--\; \; -->15)$
I did this, but I get $1.484$, whereas it's $1.038$ in my book.
I used the method:
${\log }_a<!--\; \; -->b=\frac{{\log }_c<!--\; \; -->b}{{\log }_c<!--\; \; -->a}$

Determine the following limit as x approaches 0: $\frac{\ln <!--\; \; -->(1+x)}{x}$
The process I want to take to solving this is by using the definition of the limit, but I am getting confused. ( without l'hopitals rule)
$\underset{h\to <!--\; \to \; -->0}{lim}\frac{f(x+h)-<!--\; -\; -->f(x)}{h}$
$\underset{h\to <!--\; \to \; -->0}{lim}\frac{\frac{\ln <!--\; \; -->(1+x+h)}{x+h}-<!--\; -\; -->\frac{\ln <!--\; \; -->(1+x)}{x}}{h}$
$\underset{h\to <!--\; \to \; -->0}{lim}\frac{x\ln <!--\; \; -->(1+x+h)-<!--\; -\; -->(x+h)\ln <!--\; \; -->(1+x)}{hx(x+h))}$
At this point I get confused because I know the answer is 1, but I am not getting this answer through simplification of my formula.

If $k$ is algebraically closed, we can consider a rational function as a function $k\to <!--\; \to \; -->k\cup <!--\; \cup \; -->\{+\mathrm{\infty <!--\; \infty \; -->}\}$.
Why is it important that the field be algebraically closed?

I have two questions:
1. Can a irreducible rational curve have infinitely self intersections?
2. If f has rational coefficients and the solutions for $0=f(x,y)\in <!--\; \in \; -->k[x,y]$ are parametrized by rational functions with rational coefficients of some parameter $t$, then the image of this parametrization over the rationals miss only finitely many rational points.
It is not clear to me why only finitely many points were missed. My first guess is some what related to rational points are dense. Can someone elaborate a bit geometric intuition on 2 and how to prove it(I think hint will suffice)?

List all the harmonic conjugates of the following rational function. The integration is almost impossible, Symbolab and Microsoft Math fails to arrive at the answer:
$\mathrm{\mu <!--\; \mu \; -->}(x,y)=\frac{x^2+x+y^2}{x^2+y^2}$

How to solve this natural logarithms problem?
How do I take natural logarithm of the following?
$A-<!--\; -\; -->(B{e}^{-<!--\; -\; -->xy})$

Find the domain and graph:
$f(t)=\frac{-<!--\; -\; -->t}{|t|}$
My book says to define it piecewise.
My questions:
1) Do all rational functions have to be defined piecewise, or just this one because there is an absolute value in the denominator, and the absolute value is always defined piecewise?
2) Are all rational functions defined piecewise in order to avoid having a denominator be equal to zero, is that the general reason for defining anything piecewise, to avoid having division by zero? (so then this would mean that the domain of a rational function is always defined piecewise, or only when we need to avoid having denominator be =0?)
This is how my book defines f(t) piecewise:
If $t>0$, then $|t|$ is $t$ since $t$ is already positive. For $t>0$, simplify
$f(t)=\frac{-<!--\; -\; -->t}{|t|}=\frac{-<!--\; -\; -->t}{t}=-<!--\; -\; -->1$
If $t<0$, then $|t|$ is $-<!--\; -\; -->t$ since $t$ is negative. For $t<0$, simplify
$f(t)=\frac{-<!--\; -\; -->t}{-<!--\; -\; -->|t|}=\frac{-<!--\; -\; -->t}{-<!--\; -\; -->t}=1$
(or for the last part, should the negative be inside the absolute value sign for $t<0$, as in $|-<!--\; -\; -->t|$ instead of $-<!--\; -\; -->|t|$?)
So we define it piecewise to avoid having 0 in the denominator? Because isn't absolute value defined at 0, the absolute value is continuous everywhere, and thus defined at 0?
I'm confused about defining things piecewise, and how to know when to apply a piecewise attempt in order to define a function's domain.
Also, I'm confused about the second part of this, where $t<0$ for $|t|$. How is it that here, $|t|$ is $-<!--\; -\; -->t$ if absolute value is always positive?
Maybe I'm not understanding the absolute value concept correctly, because in grade school it has always been drilled into my head that |absolute value| just "turns things positive", so here I don't really understand how it can be negative.
I understand how the domain is $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0)\cup <!--\; \cup \; -->(0,\mathrm{\infty <!--\; \infty \; -->})$, because by using open intervals we're not letting it be exactly =0, but I'm confused about the whole piecewise thing.

I noticed that for the (real) function $f(x)=\frac{1}{1-<!--\; -\; -->x}$, $f(f(f(x)=f^3(x)=x$ for all real $x$.
This surprised me, and I was naturally curious about rational functions that elicit the identity after 4 or 5 or, in general, $n$ iterations.
I looked at rational functions of the form $f(x)=\frac{1}{z-<!--\; -\; -->x}$ for $z\in <!--\; \in \; -->\mathbb{R}$ and wondered for which $z$ did $f^n(x):=f(f(...f(f(x))...))=x$.
After playing around for a bit on Wolfram and Desmos, I came up with the following conjecture: If $n\in <!--\; \in \; -->\mathbb{N}$ and $n\ge <!--\; \ge \; -->2$ and $f(x)=\frac{1}{2\cos <!--\; \; -->(\frac{\mathrm{\pi <!--\; \pi \; -->}}{n})-<!--\; -\; -->x}$, then $f^n(x)=x$.
Note that I do not claim that $f$ is the only rational function of the form $f(x)=\frac{1}{z-<!--\; -\; -->x}$ where $z\in <!--\; \in \; -->\mathbb{R}$.
So I have a few questions:
1) Is this conjecture true?
2) How would someone go about proving it, if it is true?
3) Is this a well-known theorem or a somewhat-immediate corollary or special case of a well-known theorem? If so, what is that theorem?
4) How would I find all $z\in <!--\; \in \; -->\mathbb{R}$ such that if $f(x)=\frac{1}{z-<!--\; -\; -->x}$,then $f^n(x)=x$? Is such a thing easy to do?
5) What field of math would ask questions like this?

Expressing logarithms as ratios of natural logarithms
$\frac{{\log }_2<!--\; \; -->x}{{\log }_3<!--\; \; -->x}=\frac{\ln <!--\; \; -->x}{\ln <!--\; \; -->2}÷<!--\; ÷\; -->\frac{\ln <!--\; \; -->x}{\ln <!--\; \; -->3}$
Why can logarithms be written as ratios of natural logarithms?
Can you explain it abstractly, please?
Example of an abstract explanation: the logarithm function is an isomorphism from the group of positive real numbers under multiplication to the group of real numbers under addition, represented as a function.
The teacher doesn't go into abstractions in class, so I would really like to understand it in an abstract sense. Thank you.

Adding Logarithms
Studying for my midterm.
Solve the following algebraically:
${\log }_2<!--\; \; -->x+{\log }_2<!--\; \; -->(x+4)=5$
So I know that ${\log }_b<!--\; \; -->(mn)={\log }_b<!--\; \; -->(n)+{\log }_b<!--\; \; -->(m)$ therefore:
$5={\log }_2<!--\; \; -->(x(x+4))$
$\text{or}$
$5={\log }_2<!--\; \; -->(x^2+4x)$
$5=2^{x^2+4x}$
Now I don't know how to solve for x at this point. I'm stuck. Please let me know if I've done anything wrong in my calculations. The answer key says that $x=4$

How to solve this equation with linear n as well as polynomial n?
I am banging my head against the wall, but somehow I can't find a closed form solution to this equation in n:
$229,244+58,044\cdot <!--\; \cdot \; -->n=130,000\ast <!--\; \ast \; -->{1.78}^n$
Obviously, if there was no $n$ multiplied with 58,044, then this would be trivial (take a log and solve it). I am at a loss about how to solve this one though. Any hints will be appreciated. Please don't copy paste the Wolfram Alpha solution here, as I am interested in a pen-paper solution approach to this problem.