Expert Answers to Algebra 2 Problems

$$n(\log <!--\; \; -->(n+2)-<!--\; -\; -->\log <!--\; \; -->(n))=n{\int <!--\; \int \; -->}_n^{n+2}\frac{dx}{x}={\int <!--\; \int \; -->}_0^2\frac{n}{n+x}dx$$
so, by the dominated convergence theorem,
$$\underset{n\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}}{lim}n(\log <!--\; \; -->(n+2)-<!--\; -\; -->\log <!--\; \; -->(n))={\int <!--\; \int \; -->}_0^{2}1dx=2.$$

David charges $15 plus $5.50 per hour to mow lawns. Ari charges $12 plus $6.25 per hour to mow lawns. In what situations is Ari's charge greater than or equal to David's charge?

$88,000 is borrowed for 14 years. If the interest on the loan is $115,826 22, find the monthly payment

Elementary Symmetric Means as Quasi-Arithmetic Means
The elementary symmetric polynomial of degree k in n variables is
$$e_{n,k}(\mathbf{x})=\underset{1\le <!--\; \le \; -->i_1\le <!--\; \le \; -->i_2\le <!--\; \le \; -->\cdots <!--\; \cdots \; -->\le <!--\; \le \; -->i_k\le <!--\; \le \; -->n}{\sum <!--\; \sum \; -->}{x}_{i_1}{x}_{i_2}\cdots <!--\; \cdots \; -->x_{i_k}$$
and for positive data x, the corresponding elementary symmetric mean is
$$s_{n,k}(\mathbf{x}):=\sqrt[k]{\frac{e_{n,k}(\mathbf{x})}{(\genfrac{}{}{0ex}{}{n}{k})}}.$$
Quasi-arithmetic means are defined in terms of an invertible function f, and the arithmetic mean A
$$M_f(\mathbf{x}):=(f^{-<!--\; -\; -->1}\circ <!--\; \circ \; -->A\circ <!--\; \circ \; -->f)(\mathbf{x})=f^{-<!--\; -\; -->1}\left(\frac{f(x_1)+f(x_2)+\cdots <!--\; \cdots \; -->+f(x_n)}{n}\right).$$
For any n, the elementary symmetric mean in n variables of degree n is the geometric mean, which is a quasi-arithmetic mean:
$$s_{n,n}(x_1,\dots <!--\; \dots \; -->,x_n)=\sqrt[n]{x_1{x}_2\cdots <!--\; \cdots \; -->x_n}=\exp <!--\; \; -->\left(\frac{\log <!--\; \; -->(x_1)+\log <!--\; \; -->(x_2)+\cdots <!--\; \cdots \; -->+\log <!--\; \; -->(x_n)}{n}\right).$$
My question is: are the other elementary symmetric means (with $k<n$) also quasi-arithmetic? If so, can the conjugating functions $f_{n,k}$ be described explicitly?
My attempt: Obviously $s_{n,1}=A$ is quasi-arithmetic, so the first nontrivial case is $s_{3,2}$, with the functional equation
$$f_{3,2}\circ <!--\; \circ \; -->s_{3,2}=A\circ <!--\; \circ \; -->f_{3,2},$$
or
$$f_{3,2}\left(\sqrt{\frac{x_1{x}_2+x_1{x}_3+x_2{x}_3}{3}}\right)=\frac{f_{3,2}(x_1)+f_{3,2}(x_2)+f_{3,2}(x_3)}{3}.$$
I tried pre-composing f with a logarithm and differentiation, but it didn't seem to get me anywhere.
I would greatly appreciate any help on this.

Modeling conditional probability
We are given three coins: one has heads in both faces, the second has tails in both faces, and the third has a head in one face and a tail in the other. We choose a coin at random, toss it, and the result is heads. What is the probability that the opposite face is tails?
Solution:
If $p=P(\text{Two headed coin was chosen | Heads came up})={\displaystyle \frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}}$
New to probability, would like to see how this solution works in more detail.
My questions:
1.The sample space is {HH,HT,TT}. The condition that heads came up refers to the elements {HH,HT} and the fact that two headed coin was chosen refers to {HH}. We want the probability of $\{HH,HT\}\cap <!--\; \cap \; -->\{HH\}=\{HH\}$. The probability of choosing HH is ${\displaystyle \frac{1}{3}}$ by Uniform Probability Law. Is this correct?
2. When we calculate that $P(\text{Heads came up})={\displaystyle \frac{1}{2}}$, what's the sample space? Is it {H,T} or {HH,HT,TT}?

Modeling Combinations
"We are given n points in the plane such that no three are collinear and no four lie on the same circle. How many circles are there that contain three of these points?"
Now I attempted to solve it and I ended up having 3 circles as the answer. I got there by actually drawing possibilities. However, I feel it's wrong. Any correction or explanation is very much appreciated.

Polynomials and Arithmetic: If $p(x_i)=7$ for four integers, then $p(z)\ne <!--\; \ne \; -->14$
Consider the polynomial
$$p(x)=a_0+a_{1}x+a_2{x}^2+···+a_n{x}^n$$
where $a_0,a_1,...,a_n\in \mathbb{Z}$. Show that if $p(x_i)=7$ for 4 distinct integers $x_0,x_1,x_2,x_3$, then $p(z)\ne <!--\; \ne \; -->14$ for any $z\in \mathbb{Z}$.
How do I start this question?

Why is this arbitrary-looking identity of arithmetic functions "obvious"?
The question asks us the following: if f is a multiplicative arithmetic function, and g is a completely multiplicative arithmetic function, and furthermore for all primes p and $n\ge <!--\; \ge \; -->1$ we have the relation
$$f(p^{n+1})=f(p)f(p^n)-<!--\; -\; -->g(p)f(p^{n-<!--\; -\; -->1}),$$
then for all n,m
$$f(n)f(m)=\underset{d\mid <!--\; \mid \; -->gcd(n,m)}{\sum <!--\; \sum \; -->}g(d)f\left(\frac{nm}{d^2}\right).$$
More than any exercise so far in this book, this looks completely arbitrary to me. I have solved it: you can consider the question only for $n=p^a,m=p^b$, and then it follows for arbitrary n,m by multiplicativity of f,g; and we can prove it for $p^a\le <!--\; \le \; -->p^b$ by induction on a. But that has really taught me nothing: I don't understand why it might make some sense that this is true.
If $g(p)=0$ for all primes p, then the relation we start from becomes $f(p^n)=f(p{)}^n$ - i.e., f is completely multiplicative. Thus maybe we can see the $g(p)f(p^{n-<!--\; -\; -->1})$ as a sort of error term of the complete multiplicativity of f. But this doesn't help me see why the result makes sense.
Another approach I tried was noting that the relation allows us to write $f(p^k)$ as a polynomial in f(p),g(p), but I wrote out a few and I didn't see a pattern I understood.
So my question is: where does this identity come from? How should I understand it, visualize it, "get" it? How would one come up with an exercise like this?

Solve the logarithmic equation.
$${\log }_9<!--\; \; -->x+{\log }_9<!--\; \; -->(8x-<!--\; -\; -->1)=1$$

Solving for n. Logarithm
How come:
${\displaystyle \frac{3^n}{2^{2n-<!--\; -\; -->1}}}<0.5\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->5}$
will be equal to $n=45$?

How do I solve the equation $e^{\ln <!--\; \; -->(2x+1)}=5x$?
I've tried using natural logs to both sides like.. $2x+1=\ln <!--\; \; -->5x$ But I'm not sure if $\ln$ and $e^{\ln }$ cancel out.

$a=b^x+c^x$, How to solve for $x$?
If $a=b^x$, then $x$ could be written in terms of $a$ and $b$; $x={\displaystyle \frac{\log <!--\; \; -->(a)}{\log <!--\; \; -->(b)}}$
What about $a=b^x+c^x$?
Could $x$ be written in terms of $a,b$ and $c$? $x=$?

Unsure how to treat y in this derivative/log problem
Need to find the derivative of $h(y)=\ln <!--\; \; -->(y^2\cos <!--\; \; -->y)$
Treating it like a normal variable like an x isn't working for me, the way we used y's in earlier problems where you get a y' in there doesn't seem right, so I'm not quite sure here.

Find $x={\log }_2<!--\; \; -->(4^{\frac{3}{4}}\cdot <!--\; \cdot \; -->\sqrt{2^5}{)}^{\frac{1}{2}}$
I am looking for x.
$$x={\log }_2<!--\; \; -->[(4^{\frac{3}{4}}\cdot <!--\; \cdot \; -->\sqrt{2^5}{)}^{\frac{1}{2}}]$$
I am not sure how to do this.
I am trying to solve this by changing the form like this:
$$x={\log }_a<!--\; \; -->b\Rightarrow <!--\; \Rightarrow \; -->a^x=b$$
thanks.

Need to know if the given du value for this integral/ln solution is correct
$\int <!--\; \int \; -->\frac{2}{x(\ln <!--\; \; -->x+5{)}^7}dx$ for that problem the example gives $u=\ln <!--\; \; -->x+5$ and $du=\frac{1}{x}dx$. based on a couple other problems I thought du should be $\frac{1}{x+5}dx$ The derivative of inside ln, over what's in ln left alone. Is there something I'm missing about this? Or, since it's happened quite a few times, is this written incorrectly?

Questions on a self-made theorem about polynomials
For any complex polynomial P degree n:
$$\underset{k=0}{\overset{n+1}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^k(\genfrac{}{}{0ex}{}{n+1}{k})P(a+kb)=0\mathrm{\forall <!--\; \forall \; -->}a,b\in <!--\; \in \; -->\mathbb{C}$$
Basically, if P is quadratic, $P(a)-<!--\; -\; -->3P(a+b)+3P(a+2b)-<!--\; -\; -->P(a+3b)=0$ (inputs of P are consecutive terms of any arithmetic sequence). This can be generalized to any other degrees.
- Has this been discovered? If yes, what's the formal name for this phenomenon?
- Is it significant/Are there important consequences of this being true?
- Can this be generalized to non-polynomials?

Why does $\ln <!--\; \; -->\frac{1}{2}=\ln <!--\; \; -->2$...What is the rule that makes this/ How does it come about?

Calculations on GF(16) find 0111/1111
It's my first time doing finite field arithmetics. As an exercise, I want to find $0111/1111\in <!--\; \in \; -->GF(16)$ generated by $\mathrm{\Pi <!--\; \Pi \; -->}(\mathrm{\alpha <!--\; \alpha \; -->})=1+\mathrm{\alpha <!--\; \alpha \; -->}+{\mathrm{\alpha <!--\; \alpha \; -->}}^4$ that is an irreducible polynomial.
In polynomial form we have:
$0111\to <!--\; \to \; -->\mathrm{\alpha <!--\; \alpha \; -->}+{\mathrm{\alpha <!--\; \alpha \; -->}}^2+{\mathrm{\alpha <!--\; \alpha \; -->}}^3$
$1111\to <!--\; \to \; -->1+\mathrm{\alpha <!--\; \alpha \; -->}+{\mathrm{\alpha <!--\; \alpha \; -->}}^2+{\mathrm{\alpha <!--\; \alpha \; -->}}^3$
If I perform the polynomial division, I obtain -1 (that is the same result obtained writing $0111\equiv <!--\; \equiv \; -->-<!--\; -\; -->1(\mathrm{mod}1111)$.
How can I compute this result -1 in the right element of the field? Or perhaps this some kind of sign that the result $\notin GF(16)$?

How to solve the equation $x^2({\log }_{10}<!--\; \; -->x{)}^5=100$?
Find the value of $x$ (without hit and trial) from the equation $x^2({\log }_{10}<!--\; \; -->x{)}^5=100$
Solving few steps I got $x={10}^{(\frac{x}{10}{)}^5}$

Solve:
$$\frac{1}{8}+\frac{2}{15}$$ divided by $$\frac{2}{3}$$