Expert Answers to Algebra 2 Problems

A triangle contains angles measuring 28 and 37 how many degrees is the third angle of the triangle

What is the limit of ${\log }_k<!--\; \; -->(k^a+k^b)$ for $k\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}$?
I'm not very good with analysis (I never studied it) but because of my "work" on other topics of mathematics I came to this problem.
$\underset{k\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}}{lim}{\log }_k<!--\; \; -->(k^a+k^b)=max(a,b)$
I'm sure that this is really "reasonable" because I tryed to graph it for really huge values of $k$ ... but "reasonable" is not enough in mathematics: I'd like to prve this. So I'm courious to know how one should go to prove it in a formal way ... if it is true (I hope).
PS: I know nothing about Limits and their rules

$\underset{p\le <!--\; \le \; -->x}{\sum <!--\; \sum \; -->}\frac{1}{pq}$
I was given that $\underset{p\le <!--\; \le \; -->x}{\sum <!--\; \sum \; -->}\frac{1}{p}$ = $\log <!--\; \; -->\log <!--\; \; -->x$+O(1).
I need to show that $\underset{pq\le <!--\; \le \; -->x}{\sum <!--\; \sum \; -->}\frac{1}{pq}=(\log <!--\; \; -->\log <!--\; \; -->x{)}^2+O(\log <!--\; \; -->\log <!--\; \; -->x)$
Here we go:
Break the sum into two sums: $\underset{p\le <!--\; \le \; -->x}{\sum <!--\; \sum \; -->}\frac{1}{p}\underset{q\le <!--\; \le \; -->\frac{x}{p}}{\sum <!--\; \sum \; -->}\frac{1}{q}$
Using what I was given: $(\log <!--\; \; -->\log <!--\; \; -->x+O(1))(\log <!--\; \; -->\log <!--\; \; -->\frac{x}{p}+O(1))$
Log Rules: $(\log <!--\; \; -->\log <!--\; \; -->x+O(1))(\log <!--\; \; -->(\log <!--\; \; -->x-<!--\; -\; -->\log <!--\; \; -->p)+O(1))$
Algebra: $\log <!--\; \; -->\log <!--\; \; -->x\cdot <!--\; \cdot \; -->\log <!--\; \; -->(\log <!--\; \; -->x-<!--\; -\; -->\log <!--\; \; -->p)+O(\log <!--\; \; -->\log <!--\; \; -->x)$
From here I am lost. Any ideas?

Need help with logarithmic differentitation
I have the expression
$y=\sqrt{x^2(x+1)(x+2)}.$
I have tried looking at videos but I still cannot arrive at the correct answer and don't know how to get there.
By the way, the correct answer is
$y^{\prime }=\frac{4x^2+9x+4}{2\sqrt{(x+1)(x+2)}}.$
Please, help.

Sketch the following functions using graph transformations:
$f(x)=(2x-<!--\; -\; -->1{)}^3$

Find the value of K. Use of l Hospital's rule and expansion is not allowed.
Let $f(x)={\log }_{cos3x}<!--\; \; -->(\cos <!--\; \; -->2ix)$ if $x\ne <!--\; \ne \; -->0$ and $f(0)=k$ where ($i$= iota) is continuous at $x=0$, then find the value of $K$. Use of l Hospital's rule and expansion is not allowed.

Solve $2^x=x^2$
I've been asked to solve this and I've tried a few things but I have trouble eliminating $x$. I first tried taking the natural log:
$x\ln <!--\; \; -->\left(2\right)=2\ln <!--\; \; -->\left(x\right)$
${\displaystyle \frac{\ln <!--\; \; -->\left(2\right)}{2}}={\displaystyle \frac{\ln <!--\; \; -->\left(x\right)}{x}}$
I don't know what to do from here so I decided to try another method:
$2^x=2^{{\log }_2<!--\; \; -->\left(x^2\right)}$
$x={\log }_2<!--\; \; -->\left(x^2\right)$
And then I get stuck here, I'm all out of ideas. My guess is I've overlooked something simple…

Does $\underset{n=2}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(n\ln <!--\; \; -->n{)}^{-<!--\; -\; -->1}$ diverge?
This seems like elementary calculus, but I can't figure this out. Can anyone supply a hint?

I have a hard time understanding why $\ln <!--\; \; -->e=1$
Can someone explain to me why the natural logarithm of e is exactly equal to the first nonzero but positive integer?

It is easy to show $\log$ isn't a polynomial (no continuous extension to $\mathbb{R}$). More challenging is showing it isn't rational.
Suppose it were a rational function. Then write, the fraction in lowest terms
$\log <!--\; \; -->x=\frac{G(x)}{Q(x)}⟺<!--\; ⟺\; -->\frac{G(x)}{\log <!--\; \; -->x}=Q(x)$
Clearly, as $x\to <!--\; \to \; -->0$, $Q(x)\to <!--\; \to \; -->0$. However, $Q(x)$ then has $x$ as factor so that
$\frac{G(x)}{x\log <!--\; \; -->x}=Q_2(x)$
It is well known that $x\log <!--\; \; -->x\to <!--\; \to \; -->0$ as $x\to <!--\; \to \; -->0$, so for $Q_2(x)$ to have a finite limit as $x\to <!--\; \to \; -->0$, which it must since it is a polynomial, $G(x)\to <!--\; \to \; -->0$ as $x\to <!--\; \to \; -->0$ so that $x$ is a factor of $G(x)$. This contradicts the assumption that $\frac{G}{Q}$ was in lowest terms.
If there is something wrong with this, please comment, but my main question is
What are some other ways to prove that logx isn't a rational function?

If $k$ is a field in which $1+1\ne <!--\; \ne \; -->0$, prove that $\sqrt{1-<!--\; -\; -->x^2}$ is not a rational function. Hint. Mimic the classical proof that $\sqrt{2}$ is irrational
It seems rather odd to discuss the square root function in the context of an arbitrary field, but here goes nothing!
Proof:
Suppose that $\sqrt{1-<!--\; -\; -->x^2}$ is a rational function in $k(x)$. Then there exist polynomials $p(x)$ and $q(x)\ne <!--\; \ne \; -->0$ that are relatively prime with $\sqrt{1-<!--\; -\; -->x^2}=\frac{p(x)}{q(x)}$; clearly we may take $p(x)\ne <!--\; \ne \; -->0$. Then $1-<!--\; -\; -->x^2=\frac{p(x{)}^2}{q(x{)}^2}$ or $q(x{)}^2(1-<!--\; -\; -->x^2)=p(x{)}^2$, which says $q|p^2$, and since $q|q^2$, we can infer $q|p^2$. Moreover, since $p|p^2$ and $(p,q)=1$, then $pq|p^2$ or $q|p$, which contradicts the fact that the polynomials are relatively prime. Hence $\sqrt{1-<!--\; -\; -->x^2}$ cannot be a rational function.
How does this sound? Aside from the problem of discussing –$\sqrt{}$ in the context of an arbitrary field, I am worried about not using the fact that $1+1\ne <!--\; \ne \; -->0$, at least not explicitly. Where exactly is this assumed used, if at all?
EDIT:
Suppose that $q(x)=1$, and let $f(x)=a_n{x}^n+...a_{1}x+a_0$. Then $1-<!--\; -\; -->x^2=f(x{)}^2$. If $x=0$, $1=a_0^2$ and therefore $a_0$ must be a unit. Letting $x=1$ we get $0=(a_n+...+a_1+a_0{)}^2$; and letting $x=-<!--\; -\; -->1$ we get $0=(-<!--\; -\; -->a_n-<!--\; -\; -->...-<!--\; -\; -->a_1+a_0{)}^2$. Since we are working in a field, there can be no nonzero nilpotent which means that $a_n+...+a_1+a_0$ and $-<!--\; -\; -->a_n-<!--\; -\; -->...-<!--\; -\; -->a_1+a_0$ are both zero. Adding the two equations together yields $2a_0=0$, and since $1+1\ne <!--\; \ne \; -->0$, $a_0=0$ which contradicts the fact that $a_0$ is a unit.
How does this sound?
Another attemtpt:
If $f(x{)}^2=1-<!--\; -\; -->x^2$, then $2=\mathrm{deg}<!--\; \; -->(f^2)=2\mathrm{deg}<!--\; \; -->(f)$ implies $\mathrm{deg}<!--\; \; -->(f)=1$. Yet if $x=-<!--\; -\; -->1$, we get $f(-<!--\; -\; -->1{)}^2=1-<!--\; -\; -->1=0$ and therefore $f(-<!--\; -\; -->1)=0$ since there are no nonzero nilpotent elements in a field; similarly, $f(1)=0$. Since $1+1\ne <!--\; \ne \; -->0$, $f$ has two distinct roots in $k$ yet is only a 1-st degree polynomial, a contradiction.

Something to the power of a logarithm
This is probably a very obvious question, but here goes...
An answer in my textbook claims that
$3^{\log <!--\; \; -->n}=n^{\log <!--\; \; -->3}$
and that
$4n^2(3/4{)}^{\log <!--\; \; -->n}=4n^{\log <!--\; \; -->3}$
Why, using more basic laws, is this the case?
(Unfortunately Google confuses this question with changing bases, exponentiation being the inverse of log (which is of course related), and similar matters.)

So I know that a rational function is a function that can be defined by a rational fraction and I managed to expand
$\tan <!--\; \; -->3\mathrm{\theta <!--\; \theta \; -->}=\frac{3\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->{\tan }^3<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{1-<!--\; -\; -->3{\tan }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}$
Is this considered as a rational function? Thanks.

Finding $y^{\mathrm{\prime <!--\; \prime \; -->}}$ of $y={\log }_7<!--\; \; -->e^{8x}$
I know that ${\displaystyle \frac{d}{dx}}(e^{8x})=8e^{8x}$, but I am confused on how to work the rest of the problem.
Is this correct:
${\log }_e<!--\; \; -->x=\ln <!--\; \; -->x$ and that ${\displaystyle \frac{d}{dx}}(\ln <!--\; \; -->x)={\displaystyle \frac{y^{\mathrm{\prime <!--\; \prime \; -->}}}{y}}={\displaystyle \frac{1}{x}}.$ (Can you show why ${\log }_e<!--\; \; -->x=\ln <!--\; \; -->x$ )
$y={\log }_b<!--\; \; -->(x)⟹<!--\; ⟹\; -->b^y=x⟹<!--\; ⟹\; -->\ln <!--\; \; -->(b^y)=\ln <!--\; \; -->(x)⟹<!--\; ⟹\; -->y\ln <!--\; \; -->(b)=\ln <!--\; \; -->(x)⟹<!--\; ⟹\; -->y={\displaystyle \frac{\ln <!--\; \; -->(x)}{\ln <!--\; \; -->(b)}}$
${\displaystyle \frac{d}{dx}}{\log }_b<!--\; \; -->(x)={\displaystyle \frac{1}{x\ln <!--\; \; -->(b)}}$
${\log }_a<!--\; \; -->(x^p)=p{\log }_a<!--\; \; -->x⟹<!--\; ⟹\; -->{\log }_7<!--\; \; -->(e^{8x})=8x{\log }_7<!--\; \; -->e$
$y^{\mathrm{\prime <!--\; \prime \; -->}}={\displaystyle \frac{8e^{8x}}{e^{8x}\ln <!--\; \; -->(7)}}={\displaystyle \frac{8}{\ln <!--\; \; -->(7)}}$

Find x in
$3^{{\log }_2<!--\; \; -->x}+3^{{\log }_x<!--\; \; -->2}=90$

Are rational functions between affine algebraic sets continuous with respect to the Zariski topology on their domain of definition?

Let $P$ and $Q$ be two rational functions of $z$ (coefficients over $\mathbb{C}$). How can one decide whether $z=R(P(z),Q(z))$ for some rational function $R$ of two variables, and if it is the case how can one find $R$ explicitely?
I'm interested in the particular case $P(z)=(z-<!--\; -\; -->1{)}^3/z^2$ for $c\in <!--\; \in \; -->\mathbb{C}$. I have been able to express $z$ in terms of $P$ and $Q$ using cubic and square roots, but I don't know how to proceed to find the rational expression.

Find the value of this logarithmic expression involving fifth root of unity.
Let $\mathrm{\alpha <!--\; \alpha \; -->}$ be the fifth root of unity. We then want to evaluate the expression
$\log <!--\; \; -->|1+\mathrm{\alpha <!--\; \alpha \; -->}+{\mathrm{\alpha <!--\; \alpha \; -->}}^2+{\mathrm{\alpha <!--\; \alpha \; -->}}^3-<!--\; -\; -->1/\mathrm{\alpha <!--\; \alpha \; -->}|$
Thanks in anticipation for your help in solving this!

This integral poses a challenge.
Ordinarily integrating rational functions can be solved using the Hermite-Ostrogradski method. However, in the following integral, the coefficients ${\mathrm{\beta <!--\; \beta \; -->}}_0,...,{\mathrm{\beta <!--\; \beta \; -->}}_4$ are not integers. (Hence, the Hermite-Ostrogradski method would not be appropriate).
Note: Expanding the integrand (trying to solve the integral using a partial fraction decomposition) is (because of the nature of the physical problem described by this integral) an inappropriate solution to this case.
$\int <!--\; \int \; -->{\displaystyle \frac{1}{{\mathrm{\beta <!--\; \beta \; -->}}_0+{\mathrm{\beta <!--\; \beta \; -->}}_{1}x+{\mathrm{\beta <!--\; \beta \; -->}}_2{x}^2+{\mathrm{\beta <!--\; \beta \; -->}}_3{x}^3+{\mathrm{\beta <!--\; \beta \; -->}}_4{x}^4}}dx$
How can this rational function be evaluated?