Algebra Word Problem Solutions

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.

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?

Quadratic Bézier curve points
A quadratic Bézier curve is the path traced by the function B(t), given points $P_0$, $P_1$, and $P_2$.
$C(t)=\underset{i=0}{\overset{2}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{2}{i})t^i(1-<!--\; -\; -->t{)}^{2-<!--\; -\; -->i}{P}_i$
$C(t)=(P_0-<!--\; -\; -->2P_1+P_2)t^2+(-<!--\; -\; -->2P_0+2P_1)t+P_{0}t\in <!--\; \in \; -->[0,1]$
What exactly is $P_0$ or $P_1$ or $P_2$ concerning this equation?
Yes they are points. But in my understanding, a point is a pair of numbers (in 2D-space).
Let $=(1,1)$, $P_1=(1,7)$ and $P_2=(7,1)$.
What values do you use (and where)?
How do you calculate the Bezier Curve for these points?

How do you graph y=6−1.25x?

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.

Determining quadratic function of this word problem
A tour company offers back country hiking excursions at $160 per person forgroups up to 50. For groups larger than 50, the company will reduce the cost of every ticket by $2 for each person in excess of 50. What size of group would produce the greatest revenue?
I'm not worried about finding out what the size of the group is that produces the greatest revenue. What I want to know is what the equation for this is. I was able to figure out this equation:
$f(x)=160x-<!--\; -\; -->2(x-<!--\; -\; -->50),$
where x is the number of people and f(x) is the revenue. But that's a linear function and we're doing quadratic functions. So, There should be an exponent and I'm thinking there's more to this question than I see? Or is it a linear function?

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?

How do you write the point slope form of the equation given (1,-3) and slope 0?

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.

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

Write the equation in point slope form given (4, -5), perpendicular to 2x-5y= -10

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 $81$

Find the slope that is perpendicular to the line ${\displaystyle y=\frac{7}{5}x-2}$

Write the point slope form of the equation given (-4,0) parallel to ${\displaystyle y=\frac{3}{4}x-2}$

Ben was walking in Manhattan at a constant speed. He went along 5th avenue from 4th to 98th street. At 3:00 he was on 15th street, and at 4:30 he was on 45th street. When did he start and when did he finish his walk?

Write the rule for the nth term given 1,10,100,1000,...

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?

How is discriminant related to real x?
Solve for range of the function,
$y=\frac{x^2+4x-<!--\; -\; -->1}{3x^2+12x+20}$
Text book says, cross multiply and express the obtained equation as a quadratic equation in x
So I get $(3y-<!--\; -\; -->1)x^2+(12y-<!--\; -\; -->4)x+(20y-<!--\; -\; -->1)=0$
Now it says find discriminant D, so we have,
$D=-<!--\; -\; -->4(3y-<!--\; -\; -->1)(8y+3)$
Now it says, set $D\ge 0$ as x is real. Wait what?
Isn't $x\in <!--\; \in \; -->\mathbb{R}$ the domain for a quadratic function? Meaning "x" is always real? What does a discriminant got anything to do with x being real, when all discriminant tells us is whether or not the ROOTS are real? Help please.
To be more specific about my doubt, here's an edit.