Expert Answers to Algebra 2 Problems

Why is the triangle inequality equivalent to $a^4+b^4+c^4\le <!--\; \le \; -->2(a^2{b}^2+b^2{c}^2+c^2{a}^2)$
Consider the existential problem of a triangle with side lengths $a,b,c\ge <!--\; \ge \; -->0$. Such a triangle exists if and only if the three triangle inequalities
$\begin{array}{}\text{(0)}& a+b\ge <!--\; \ge \; -->c,b+c\ge <!--\; \ge \; -->a\text{and}c+a\ge <!--\; \ge \; -->b\end{array}$
are all satisfied.
Alternatively, if ${\mathrm{\ell <!--\; \ell \; -->}}_1\le <!--\; \le \; -->{\mathrm{\ell <!--\; \ell \; -->}}_2\le <!--\; \le \; -->{\mathrm{\ell <!--\; \ell \; -->}}_3$ are the values of a,b and c ordered in ascending order, then the triangle exists iff ${\mathrm{\ell <!--\; \ell \; -->}}_1+{\mathrm{\ell <!--\; \ell \; -->}}_2\ge <!--\; \ge \; -->{\mathrm{\ell <!--\; \ell \; -->}}_3$.
Interestingly, the three triangle inequalities can be recast into a single quartic polynomial inequality. Let $0,x,y\in <!--\; \in \; -->{\mathbb{R}}^2$ be the three vertices of the triangle, with $\Vert <!--\; \Vert \; -->x\Vert <!--\; \Vert \; -->=a,\Vert <!--\; \Vert \; -->y\Vert <!--\; \Vert \; -->=b$ and $\Vert <!--\; \Vert \; -->x-<!--\; -\; -->y\Vert <!--\; \Vert \; -->=c$. Then $c^2=\Vert <!--\; \Vert \; -->x-<!--\; -\; -->y{\Vert <!--\; \Vert \; -->}^2=\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}^2-<!--\; -\; -->2⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->+\Vert <!--\; \Vert \; -->y{\Vert <!--\; \Vert \; -->}^2=a^2+b^2-<!--\; -\; -->2⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->$
Therefore $x^{T}y=⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->=\frac{1}{2}(a^2+b^2-<!--\; -\; -->c^2)$ and
$\begin{array}{}\text{(1)}& \left(\begin{array}{c}{x}^T\\ y^T\end{array}\right)\left(\begin{array}{cc}x& y\end{array}\right)=\frac{1}{2}\left(\begin{array}{cc}2a^2& a^2+b^2-<!--\; -\; -->c^2\\ a^2+b^2-<!--\; -\; -->c^2& 2b^2\end{array}\right).\end{array}$
The RHS of (1) must be positive semidefinite because the LHS is a Gram matrix. Conversely, if the RHS is indeed PSD, it can be expressed as a Gram matrix. Hence we obtain x and y and the triangle exists.
As $2a^2$ and $2b^2$ are already nonnegative, the RHS of (1) is positive semidefinite if and only if $(2a^2)(2b^2)-<!--\; -\; -->(a^2+b^2-<!--\; -\; -->c^2{)}^2\ge <!--\; \ge \; -->0$, by Sylvester's criterion. That is, the triangle exists if and only if
$\begin{array}{}\text{(2)}& -<!--\; -\; -->(a^4+b^4+c^4)+2(a^2{b}^2+b^2{c}^2+c^2{a}^2)\ge <!--\; \ge \; -->0.\end{array}$
This polynomial inequality can be derived by more elementary means. See circle-circle intersection on Wolfram MathWorld. The geometric explanation for the necessity of (2) is given by Heron's formula, which states that the square root of the LHS is four times the area of the triangle.
Since both (0) and (2) are necessary and sufficient conditions for the existence of the required triangle, the two sets of conditions must be equivalent to each other. Here are my questions. Is there any simple way to see why (0) and (2) are equivalent? Can we derive one from the other by some basic algebraic/arithmetic manipulations?

Solve the equations
$x+y=35$
$x-<!--\; -\; -->y=13$

Evaluate using long division first to write f(x) as the sum of a polynomial and a proper rational function.
$\int <!--\; \int \; -->\frac{xdx}{3x-<!--\; -\; -->4}$

Inverting the exponential decay equation?
How would I algebraically isolate $x$ in the following exponential decay equation, so that $x$ is most easily derived?
$y=A{e}^{-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}x}$

Solving ${\log }_2<!--\; \; -->(x^2)=x$ explicitly?
I'm having problems getting a proper step-by-step solution to this equation.
${\log }_2<!--\; \; -->(x^2)=x$
I know the results are 2 and 4, but so far I can get only solutions like these:
$2^x=x^2\text{or}\frac{\log <!--\; \; -->(x)}{x}=\frac{\log <!--\; \; -->(2)}{2}.$
I'd like something along the lines of $x=\cdots <!--\; \cdots \; -->$
Is it possible with some basic knowledge?

Derivative of log and exponential function
How would you find the derivative of:
$y=e^{\sqrt{x}}+\ln <!--\; \; -->\sqrt{x}$
Would I simply use the product rule?

Algebra: logarithms word question
Initially there are $2000$ bacteria in a given culture. The number of bacteria $N$ is tripling every hour so $N=2000\cdot <!--\; \cdot \; -->3^t$, where $t$ is the measurable in hours.
a) How many bacteria are present after 4 hours?
b) How long is it until there are 1000000 bacteria?

What $n^{\frac{1}{{\log }_2<!--\; \; -->n}}$ means?
I was confused with about the $n^{\frac{1}{{\log }_2<!--\; \; -->n}}$ expression. I am not sure how to make mathematical sense of it - i.e. express it in another way for easier understanding. I tried to plug in some numbers like 4 and 8. I got the answer 2 in both cases. Is it what this expression means? 2?

Jim wants to build a rectangular parking lot along a busy street but only has 1900 feet of fencing available. If no fencing is required along the street, find the maximum area of the parking lot.

Do all monic polynomials factor into monic irreducibles?
Do all monic polynomials in $\mathbb{Z}/n\mathbb{Z}[x]$ where n is prime factor into monic irreducibles?
Just by thinking about these monic polynomials and how they only act on a finite number of elements makes me think that the assertion is true, because if d is a root of a monic polynomial f(x) then $f(x)=(x-<!--\; -\; -->d)g(x)$ where g(x) is monic, and you could apply the same rule iteratively until you exhaust all roots. But I am not sure how to really prove this, it sounds really similar to one part of the fundamental theorem of arithmetic, but I can't find how to quantify a polynomial the same way you would an integer for an inductive proof. How would I go about starting a proof of this?

Proving a logarithmic inequality
I'm interested why this is true:
$\text{Considering }\mathrm{\forall <!--\; \forall \; -->}(x,y,z)\in <!--\; \in \; -->(1,\mathrm{\infty <!--\; \infty \; -->})$
The following holds:
${\log }_x<!--\; \; -->y^z+{\log }_x<!--\; \; -->z^y+lo{g}_y{z}^x\ge <!--\; \ge \; -->\frac{3}{2}$
This is taken from a high school textbook of mine. I tried finding a meaningful manipulation by using AM-GM, but that got pretty messy. I'd like to avoid Lagrange multipliers since this is meant to be a pretty basic problem.
Any progress would be appreciated.

Modeling some constraints
We have two decision variables $x\in <!--\; \in \; -->{\mathbb{Z}}^{0+}$ that is the main decision variable and $0⩽<!--\; ⩽\; -->y⩽<!--\; ⩽\; -->1$ that is an auxiliary decision variable.
Now based on the nature of the problem we are studying, we know that
$$\begin{gathered} y=\frac{1}{x}, \\ x=0\iff <!--\; \iff \; -->y=\mathrm{\varnothing <!--\; \varnothing \; -->}, \\ x>0\iff <!--\; \iff \; -->y⩾<!--\; ⩾\; -->0. \end{gathered}$$
Now the challenge is how to formulate the dynamics of these two decision variables in the constraints of a mathematical program. We have x in the denominator and since at optimality x can actually be zero, this would create a problem. Obviously, we cannot write $xy=1$ because if the optimal solution is that $x=0$, the equality would be violated.
Any suggestions would be much appreciated.

Modular arithmetic with polynomial
Given $n=pq$ where $p\mathrm{\&<!--\; \&\; -->}q,p\ne <!--\; \ne \; -->q$ are primes, P(x) is polynomial and $z\in <!--\; \in \; -->{\mathbb{Z}}_n$
I need to prove that:
$p(z)\equiv <!--\; \equiv \; -->0(\mathrm{mod}n)⟺<!--\; ⟺\; -->p(z(\mathrm{mod}p))\equiv <!--\; \equiv \; -->0(\mathrm{mod}p)\wedge <!--\; \wedge \; -->p(z(\mathrm{mod}q))\equiv <!--\; \equiv \; -->0(\mathrm{mod}q)$
If i could prove the more general case: $P(z)\mathrm{mod}n\equiv <!--\; \equiv \; -->P(z\mathrm{mod}p)\mathrm{mod}p$ (q is the same) then i could also prove what i need of course. from my understanding so far, the above is true, but i can't figure out a way to prove this.
Back to the original question, i tried to see why the fact that p divides P(z mod p) and q divides P(z mod q) implies that pq divides P(z) and vice versa, but i didn't have much success with this.

Prove that given projective curve has genus 1.
Show that
$X=\{((x_0:x_1),(y_0:y_1)):(x_0^2+x_1^2)(y_0^2+y_1^2)=x_0{x}_1{y}_0{y}_1)\}\subseteq <!--\; \subseteq \; -->{\mathbb{P}}^1\times <!--\; \times \; -->{\mathbb{P}}^{\mathbb{1}}$
is a smooth curve of genus 1.
I can prove it with the following reasoning.
Using the Segre Embedding, the curve consists on all the elements $(x:y:z:w)\in <!--\; \in \; -->{\mathbb{P}}^3$ satisfying the equations
$x^2+y^2+z^2+w^2-<!--\; -\; -->xw=0,xw-<!--\; -\; -->yz=0,$
i.e. the vanishing set of I=$I=(x^2+y^2+z^2+w^2-<!--\; -\; -->xw,xw-<!--\; -\; -->yz)$. I can use the Jacobian criterion to prove it's smooth; no issue.
To find the genus, since I'm better at it, I decided to compute the arithmetic genus. To do that, I proved that
$\{x^2+y^2+z^2+w^2-<!--\; -\; -->xw,xw-<!--\; -\; -->yz\}$
is in fact a Gröbner basis (Using GRevLex ordering), so
$LT(I)=(x^2,yz).$
Then every minimal free resolution of the quotient over a monomial ideal (I think I can remove the hypothesis of it being monomial; I'm not completely sure, but in such case I don't need to compute LT(I) or even prove that I have a Gröbner basis) generated by two elements has the form $0\to <!--\; \to \; -->S\to <!--\; \to \; -->S^2\to <!--\; \to \; -->S\to <!--\; \to \; -->0$, in particular, in this case it has the form
$0\to <!--\; \to \; -->S(-<!--\; -\; -->4)\to <!--\; \to \; -->S(-<!--\; -\; -->2{)}^2\to <!--\; \to \; -->S\to <!--\; \to \; -->0$
which allows me to compute the Hilbert polynomial of X, by the method in the first section of The Geometry of Syzygies from Eisenbud (I remember it's also used in Cox's), and with the Hilbert polynomial I also have that the arithmetic genus is 1.
But can I prove it without computing the Hilbert polynomial?

Ambiguity with Logs and Inequalities
Solve the following inequality:
${0.8}^x>0.4$
Method 1 (Using the Common Logarithm)
${\log }_{10}<!--\; \; -->{0.8}^x>{\log }_{10}<!--\; \; -->0.4$
$x{\log }_{10}<!--\; \; -->0.8>{\log }_{10}<!--\; \; -->0.4$
Because
${\log }_{10}<!--\; \; -->0.8<0$
The sign of the inequality will change when we divide both sides of the equation by the LHS of the above inequality:
$x<\frac{{\log }_{10}<!--\; \; -->0.4}{{\log }_{10}<!--\; \; -->0.8}$
Using a reverse of the change of base theorem, we get:
$x<{\log }_{0.8}<!--\; \; -->0.4$
Method 2 (Using another based Logarithm)
${\log }_{0.8}<!--\; \; -->{0.8}^x>{\log }_{0.8}<!--\; \; -->0.4$
$x{\log }_{0.8}<!--\; \; -->0.8>{\log }_{0.8}<!--\; \; -->0.4$
Because
${\log }_{0.8}<!--\; \; -->0.8>0$
The sign of the inequality will not change when we divide both sides of the equation by the LHS of the above inequality:
$x>\frac{{\log }_{0.8}<!--\; \; -->0.4}{{\log }_{0.8}<!--\; \; -->0.8}$
The denominator of the RHS of the inequality is equal to one, therefore:
$x>{\log }_{0.8}<!--\; \; -->0.4$
The sign is reversed using both methods - the textbook states the answer to be the one gotten using the first method, but I can't spot where I went wrong in method two.
Any help will be greatly appreciated, thanks in advance.

Solve ${\log }_2<!--\; \; -->x={\log }_4<!--\; \; -->(x+6)$ for $x$ using the change of base formula.
I already tried changing the base on both sides but that didn't work I know it must be in the form of a quadratic for a substitution to be made.

What is the largest $n$-digit number which is also an exact $n$th power?
For example,the largest $2$-digit number which is an exact $2$nd power is $81$

Fine the equation of the tangent line to the curve at the point (0, 4)
$4+6y=\sin <!--\; \; -->(x{y}^2)+y$

Show that $g(x)=x\ln <!--\; \; -->x$ and $g(x)=e^x$ are bounded below.
Show that $g(x)$ is bounded below, for $0\le <!--\; \le \; -->x$
a) $g(x)=\{\begin{array}{ll}0& \text{if }x=0\\ x\ln <!--\; \; -->x& \text{if }x>0\end{array}$
b) $g(x)=e^x$
For (a), $g(x)\ge <!--\; \ge \; -->0$ for $x\ge <!--\; \ge \; -->0$
For (b), $g(x)\ge <!--\; \ge \; -->1$ for $x\ge <!--\; \ge \; -->0$
Is that all I have to say? Or is there a more technical definition/proof?

Which expression is equivalent to $\frac{42x^{-<!--\; -\; -->4}{y}^{10}{z}^3}{6x{y}^{-<!--\; -\; -->2_Z}}$