Expert Answers to Algebra 2 Problems

logarithms and function
If ${\log }_2<!--\; \; -->(f(x)+|\sin <!--\; \; -->x|)={\log }_2<!--\; \; -->x$ then:
A) $f(x)>0$ for each $x\in <!--\; \in \; -->R$
B) $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(x)=+\mathrm{\infty <!--\; \infty \; -->}$
C) the function is strictly increasing
D) $f(\mathrm{\pi <!--\; \pi \; -->})=\mathrm{\pi <!--\; \pi \; -->}$
So firstly I define domain $f(x)+|\sin <!--\; \; -->x|>0⟹<!--\; ⟹\; -->f(x)>-<!--\; -\; -->|\sin <!--\; \; -->x|$
And we have $f(x)+|\sin <!--\; \; -->x|=x⟹<!--\; ⟹\; -->f(x)=x-<!--\; -\; -->|\sin <!--\; \; -->x|$ but I'm not sure about next step
$x-<!--\; -\; -->|\sin <!--\; \; -->x|>-<!--\; -\; -->|\sin <!--\; \; -->x|⟹<!--\; ⟹\; -->x>0$
So the answers B, C, D are correct?

What is the number of real roots of $(\log <!--\; \; -->x{)}^2-<!--\; -\; -->\lfloor <!--\; \lfloor \; -->\log <!--\; \; -->x\rfloor <!--\; \rfloor \; -->-<!--\; -\; -->2=0$ $\lfloor <!--\; \lfloor \; -->\cdot <!--\; \cdot \; -->\rfloor <!--\; \rfloor \; -->$represents the greatest integer function less than or equal to x.
I know how to solve logarithm equation but due to greatest integer function I am unable to proceed further please help thanks.

Find the value of $x$ such that $2^x=10$

Growth rate of $1/(\log <!--\; \; -->(x)-<!--\; -\; -->\log <!--\; \; -->(x-<!--\; -\; -->1))$
Let $x>1$ be a real number. Let $y={\displaystyle \frac{1}{\log <!--\; \; -->(x)-<!--\; -\; -->\log <!--\; \; -->(x-<!--\; -\; -->1)}}$
My question: Approximately how fast does $y$ grow (asymptotically) in terms of $x$? (e.g. linear, polynomial, exponential)?

Is $O(n\log <!--\; \; -->n)$ always smaller than $O(m)$ for $n-<!--\; -\; -->1<m<n^2$?
I am writing an algorithm that needs to finish in $O(m)$. The problem is for a graph $G(V,E)$, where $m=|E|$ and $n=|V|$. $m$ can be in the range of $n-<!--\; -\; -->1$ to $n^2-<!--\; -\; -->1$
If I do some processing that takes $O((n-<!--\; -\; -->3)\log <!--\; \; -->(n-<!--\; -\; -->3))$ time, will that still satisfy being smaller than $O(m)$ or not?
Thanks

Prove that ${\log }^2<!--\; \; -->5+{\log }^2<!--\; \; -->7>\log <!--\; \; -->12$
What I tried so far:
${\log }^2<!--\; \; -->5+{\log }^2<!--\; \; -->7>\log <!--\; \; -->3+\log <!--\; \; -->4$
$(\log <!--\; \; -->5+\log <!--\; \; -->7{)}^2-<!--\; -\; -->2\cdot <!--\; \cdot \; -->\log <!--\; \; -->5\cdot <!--\; \cdot \; -->\log <!--\; \; -->7>\log <!--\; \; -->3+\log <!--\; \; -->4$
But it seems that I'm not even near the result.
Every suggestion / hint would be appreciated :)
Thanks in advance.
EDIT: $\log$ means ${\log }_{10}$

Diffie hellman and the discrete algorithm problem
Suppose Alice and Bob are exchanging keys using Diffie-Hellman Key-Exchange Algorithm.
a - Alice secret key
g - generator
p - prime
x - the public key passed from Alice to Bob.
Eve is listening to the communication and she is exposed to the three parameters g,p,x.
She's using a brute-force method to find $a$, Alice's secret key.
Thus, Eve is looking for a satisfying this equation:
$g^{a}modp=x$
Now, I know (By testing) Eve can find $a^{\prime }\ne <!--\; \ne \; -->a$ satisfying the equation above, and by using $a^{\prime }$ she can also compute the common secret key used by Alice and Bob.
Why is it mathematically true?

Let $X$ be an irreducible, normal variety over an algebraically closed field of characteristic zero. Let $x,y\in <!--\; \in \; -->X$ be two points such that $f(x)=f(y)$ for every $f\in <!--\; \in \; -->K(X)$ which is defined at $x$ and at $y$. Can I conclude that $x=y$?
I feel the answer should be affirmative. In fact, the statement can be reduced to the following: Given two effective prime divisors $D_x$ and $D_y$ on $X$, there exists a rational function $f\in <!--\; \in \; -->K(X)$ with $v_{D_x}(f)\ne <!--\; \ne \; -->0$ and $v_{D_y}(f)=0$.
If this is true, then assuming $x\ne <!--\; \ne \; -->y$ we could find a divisor $D_x$ containing $x$ but not $y$ and a divisor $D_y$ containing $y$ but not $x$, so a function $f$ as above would yield a contradiction. However, I just can't prove the statement, even though I also think it should be true.

I have to construct a rational function with the range being [-1,0) , which is pretty much just -1. I came up with the solution $\sqrt{-<!--\; -\; -->x^2-<!--\; -\; -->\frac{1}{x}}$. It works for the range, but I'm not sure if it is a rational function.

Next number that is a 1 and zeroes
There's probably a much better way of expressing this, but I don't know it, so I guess that's part of the question too. I'm not even sure what to tag it.
How do I find the next number greater than a given number that is a 1 followed by zeroes.
For instance
12 => 100
9343 => 10,000
etc.
Thanks

How to find the power series of rational functions of type, e.g.
$\frac{1}{1-<!--\; -\; -->6x+12x^2},\frac{6x}{1-<!--\; -\; -->6x+12x^2}$
where denominator can't be factorized over the real domain.
Is there a way to use the method of undetermined coefficients (by completing the square), or is it necessary to use complex domain and trigonometry?

I'm learning about holes in rational functions in precalculus, and I'm confused as to why they exist, having some knowledge of limits. Say I have a function $R(x)=\frac{(x-<!--\; -\; -->5)(x+2)}{(x+4)(x+2)}$. If I were to find where the hole lies for this function, I would plug −2 into $\frac{x-<!--\; -\; -->5}{x+4}$ which would give me a hole at $(-<!--\; -\; -->2,-<!--\; -\; -->\frac{7}{2})$. However, that turns out to be the limit of $R(x)$ (using L'Hopital's Rule):
$\underset{x\to <!--\; \to \; -->-<!--\; -\; -->2}{lim}\frac{(x-<!--\; -\; -->5)(x+2)}{(x+4)(x+2)}=\underset{x\to <!--\; \to \; -->-<!--\; -\; -->2}{lim}\frac{x^2-<!--\; -\; -->3x-<!--\; -\; -->10}{x^2+6x+8}=\underset{x\to <!--\; \to \; -->-<!--\; -\; -->2}{lim}\frac{2x-<!--\; -\; -->3}{2x+6}=-<!--\; -\; -->\frac{7}{2}$
So why is there a hole in the graph if the limit exists at $(-<!--\; -\; -->2,-<!--\; -\; -->\frac{7}{2})$? Am I misunderstanding the definition of a limit? Thanks.

Let, $\frac{P(x)}{Q(x)}$ and $\frac{M(x)}{N(x)}$ be two rational functions consisting of polynominals $P(x),Q(x),M(x),N(x)$. The polyniminals are defined with respect to variable x. Now, if a rational function is formed by summation of the rational functions, does there exist a general way to evaluate the zeros of $\frac{P(x)}{Q(x)}+\frac{M(x)}{N(x)}$? If yes, where will the zeros of $\frac{P(x)}{Q(x)}$ and $\frac{M(x)}{N(x)}$ map to?
Edit 1: I know that, roots of $P(x)N(x)$ will be union of the roots of P(x) and N(x), and so on so forth for $Q(x)M(x)$. So, few possible roots can be calculated by taking the intersection of the roots of $P(x)N(x)$ and $Q(x)M(x)$. But I am not interested in them, I want solution in general.
Edit 2: What if, $\frac{P(x)}{Q(x)}$ is a Pade approximant of order (m,n)?

I have a function $f(a,b)=\frac{ab}{(\frac{a+b}{2}{)}^2}$, and (to me) it has some cool properties (e.g $f(a,b)=f(b,a)$, $f(x,0)=0$, $f(x,x)=1$, $0\le <!--\; \le \; -->f\le <!--\; \le \; -->1$, etc.). Now I wanted to know the inverse function for this. It's a rational function, but I couldn't figure out how to do this.
$$\begin{gathered} y=\frac{xb}{(\frac{x+b}{2}{)}^2}=\frac{4xb}{(x+b{)}^2} \\ x=\frac{4yb}{(y+b{)}^2}⟹<!--\; ⟹\; -->x(y+b{)}^2=4yb⟹<!--\; ⟹\; -->\frac{(y+b{)}^2}{y}=\frac{4b}{x} \\ \frac{y^2+b^2}{y}=\frac{4b}{x}-<!--\; -\; -->2b⟹<!--\; ⟹\; -->y+\frac{b^2}{y}=\frac{4b}{x}-<!--\; -\; -->2b \end{gathered}$$
At this point, I have no idea where to go. I think that I'm missing out on some fundamental rule of solving this, but I don't have a clue about where to go. How do I simplify the $y+\frac{b^2}{y}$?
NOTE: The final function $f^{-<!--\; -\; -->1}(y,b)$ is such that $f^{-<!--\; -\; -->1}(f(x,b),b)=x$ for any $x$ and $b$ such that $x$ and $b$ are not both 0. One can be zero, but not both (since $f(0,0)$ is undefined).

find the domain of root of a logarithmic function
I'm a little confused about this question since output of a logarithmic function varies from $-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$ to $\mathrm{\infty <!--\; \infty \; -->}$ .I should find the domain of this function: $y=\sqrt{{\log }_x<!--\; \; -->(10-<!--\; -\; -->x^2)}$ . How can I find the interval that makes ${\log }_x<!--\; \; -->(10-<!--\; -\; -->x^2)$ greater than zero?

${\log }_{-<!--\; -\; -->3}<!--\; \; -->\left(9\right)$
Can you have a negative base for a logarithm? If the answer is no, why can you not? Isn't the answer to the above just $2$?
Similarly, is $-<!--\; -\; -->3$ also a solution to the equation ${\log }_x<!--\; \; -->\left(9\right)=2$ (I know one solution is obviously 3).

Please explain this simple rule of logarithms to me
Right I know this one is simple and I know that I just need a push to make it sink in in my head..
I am studying control systems and in one of the tutorial examples the tutor says
Show that
$20\log <!--\; \; -->(1/x)=-<!--\; -\; -->20\log <!--\; \; -->(x)$
I know that when you have a divide or a multiply with logarithms you add them and subtract them but for my own understanding I just need someone to like slowly show me how this works..
If I take the log of the numerator I have $20\log <!--\; \; -->(1)=0$ but I don't know where to go from here.. So do I now just take the log of the denominator and as the numerator was zero it is just minus whatever the log of the denominator is... Getting myself a bit muddled.. Thanks

$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}n\sqrt{2}(\sqrt{\ln <!--\; \; -->(n+1)}-<!--\; -\; -->\sqrt{\ln <!--\; \; -->n})=0$
But I haven't any ideas how to do it... My calculations shows that this sequence is monotonously decreasing.
I've proved that using inequality $\ln <!--\; \; -->(1+x^a)\le <!--\; \le \; -->ax$ and double-sided theorem.