Expert Answers to Algebra 2 Problems

Confusion about genus-degree formula
My question regards a perplexity I have on how to apply the genus-degree formula for irreducible, projective, complex plane curves. Consider first the affine complex plane curve given by the equation
$$C:(x-<!--\; -\; -->2)(x-<!--\; -\; -->1)(x+1)(x+2)-<!--\; -\; -->y^2=x^4-<!--\; -\; -->5x^2+4-<!--\; -\; -->y^2=0.$$
The Jacobian is given by $(x(2x+\sqrt{10})(2x-<!--\; -\; -->\sqrt{10}),-<!--\; -\; -->2y)$, so it never vanishes on C. Let us now look at the compactification $\stackrel{^<!--\; ^\; -->}{C}$, which has the same points as C plus a point at infinity with coordinates $[x:y:z]=[0:1:0]$. This point lies in the affine chart with $y=1$, and the affine equation for $\stackrel{^<!--\; ^\; -->}{C}$ in terms of x and z in that chart is
$$x^4-<!--\; -\; -->5z^2{x}^2+4z^4-<!--\; -\; -->z^2=0.$$
The differential $(4x^3-<!--\; -\; -->10x{z}^2,-<!--\; -\; -->10x^{2}z+16z^3-<!--\; -\; -->2z)$ vanishes at $(x,z)=(0,0)$, hence $\stackrel{^<!--\; ^\; -->}{C}$ has one singularity: the point at infinity. If we pretend for a moment that it doesn't (i.e., that it is smooth), one can do the "usual construction" to see that it is topologically a torus: one can draw two cuts along [-2,-1] and [1,2] on two copies of the Riemann sphere and glue them together with the right orientation. So if $\stackrel{^<!--\; ^\; -->}{C}$ were regular, it would be an elliptic curve, and in particular have genus 1. However, the genus-degree formula for projective plane curves predicts genus 3, since the equation of $\stackrel{^<!--\; ^\; -->}{C}$ has degree 4. But the Wikipedia article on the genus-degree formula also mentions that the formula actually gives the arithmetic genus and that for every ordinary singularity of multiplicity r, the geometric genus is smaller than the arithmetic genus by $\frac{1}{2}r(r-<!--\; -\; -->1)$. Now, I am not really sure about how to measure the multiplicity of a singularity, but in this case it seems that for any value of $r\ge <!--\; \ge \; -->0$, we never have that $3-<!--\; -\; -->\frac{1}{2}r(r-<!--\; -\; -->1)=1$. So the geometric intuition and the formula seem to disagree. The only other thing that comes to my mind is that I have not checked yet that $\stackrel{^<!--\; ^\; -->}{C}$ is irreducible, but this can be checked on C using Eisentein's criterion applied to the polynomial ring $(\mathbb{C}[x])[y]$ using the prime ideal $\mathfrak{p}=(x+1)$.
Reassuming, my question is: what is the genus of $\stackrel{^<!--\; ^\; -->}{C}$? If it is 1, why is the genus-degree formula wrong? If it is 3, why is the geometric intuition wrong? After all, also the article of Wikipedia on elliptic curves seems to confirm that $\stackrel{^<!--\; ^\; -->}{C}$ should have genus 1.

Modeling the maximum number of moves in Tower of Hanoi problem
What would be the recursive algorithm for solving the Tower of Hanoi problem (with n disks and 3 pegs) in maximal number of moves (i.e. going through all possible disks/pegs combinations).

How to calculate multiplicative inverses in $GF(2^3)$ without the Euclidean algorithm
The problem I have concerns finite field arithmetic in $GF(p^k)$
I know how to find multiplicative inverses using the extended Euclidean algorithm, but for my exams I need to calculate multiplicative inverses in $GF(2^3)$ without it.
What's the best way to do this? Is there even a convenient way?
The irreducible polynomial I have is $x^3+x+1$.

Logarithmic system of equations
Solve these equations
$$\log <!--\; \; -->x+\frac{\log <!--\; \; -->x+8\log <!--\; \; -->y}{{\log }^2<!--\; \; -->x+{\log }^2<!--\; \; -->y}=2$$
$$\log <!--\; \; -->y+\frac{8\log <!--\; \; -->x-<!--\; -\; -->\log <!--\; \; -->y}{{\log }^2<!--\; \; -->x+{\log }^2<!--\; \; -->y}=0$$

Intutively, why does $x^{\frac{1}{\ln <!--\; \; -->x}}=e$?
$x><!--\; >\; -->0$, we have this identity:
$$x^{\frac{1}{\ln <!--\; \; -->x}}=e\text{.}$$
You can see this by using the fact that $x=e^{\ln <!--\; \; -->x}$
I'm wondering if there's a good intuitive explanation for this one, given that $x^{\frac{1}{k}}$ is the operation that inverts raising $x$ to the $k$th power and $\ln <!--\; \; -->x$ is the inverse of the exponential function. Is there some compelling intuitive or geometric argument that makes this identity more obvious than algebraic rearrangement?

Solving logarithmic equations including x
Let
$${\log }_3<!--\; \; -->(x-<!--\; -\; -->2)=6-<!--\; -\; -->x$$
It's obvious drawing the graphs of the two functions that the only solution is $x=5$. But this is not really a proof, rather than observation.
How do you prove it algebraically?

Logarithmic Differentiation equation, Help!
So, I have to differentiate this via $\log$. I am still learning, so please be patient, I will try to explain everything I did. Please tell me if it is correct.
$$y=\frac{(x+3{)}^4(2x^2+5x{)}^3}{\sqrt{4x-<!--\; -\; -->3}}$$
$$\ln <!--\; \; -->(y)=\ln <!--\; \; -->((x+3{)}^4(2x^2+5x{)}^3))-<!--\; -\; -->\ln <!--\; \; -->\left(\sqrt{4x-<!--\; -\; -->3}\right)$$
$$\ln <!--\; \; -->(y)=4\ln <!--\; \; -->(x+3)3\ln <!--\; \; -->(2x^2+5x)-<!--\; -\; -->\frac{1}{2}\ln <!--\; \; -->(4x-<!--\; -\; -->3)$$
aaaaaaaaand don't know what to do next, any help in the process or next step?

Some search on the internet and this site didn't result in any topic about this question of Silverman's The Arithmetic of Elliptic Curves:
Let $W\subset <!--\; \subset \; -->{\mathbb{P}}^{\mathbb{n}}$ be a smooth algebraic set, each of whose component varieties has dimension $n-<!--\; -\; -->1$. Prove that W is a variety.
Hint: use Krull's Hauptidealsatz to show W is the zeroset of a single homogenous polynomial.
my ideas: So if I(V) contains at least two linear independent polynomials, we have to prove the height of I(V) is at least 2.
After that, it is enough to find one singular point, where we assume $I(V)=fg$ with f,g nonconstant polynomials.

One Chinese Jin is equal to 500 grams. On earth, one kilogram weighs approximately 2.2 pounds. My husband weight 145 pounds. How many Jin is his mass?

Consider a post office with two clerks. John, Paul, and Naomi enter simultaneously. John and Paul go directly to the clerks, while Naomi must wait until either John or Paul is finished before she begins service.
(a) If all of the service times are independent exponentially distributed random variables with the same mean 1/lambda, what is the probability that Naomi is still in the post office after the other two have left?
What if we change the assumption in (a) that the two clerks have different service rates lambda_1 and lambda_2?

What is the difference between the largest and smallest possible positive roots?
I am faced with the following question:
What is the difference between the largest and the smallest possible positive roots of 4x5+3x3−5x2+7x−12?
Now, my first attempt was to try substituting arbirtrary values to find one root and then long division to find the others. However, no integer (or fractional) value seemed to satisfy this.
Is the another way to approach this problem, or am I just making a simple arithmetic mistake?
Any help will be appreciated.

Solve logarithmic equation $3^{{\log }_3^2<!--\; \; -->x}+x^{{\log }_3<!--\; \; -->x}=162$
Find $x$ from logarithmic equation
$$3^{{\log }_3^2<!--\; \; -->x}+x^{{\log }_3<!--\; \; -->x}=162$$
I tried solving this, with basic logarithmic laws, changing base, etc., but with no result, then I went to wolframalpha and it says that its alternate form is:
$$2e^{\frac{{\log }^2<!--\; \; -->x}{\log <!--\; \; -->3}}=162$$
But I don't know how it came to this result, can you help me guys?

Expanding logarithm of function
Is there a way (there has to be), I can expand an expression like this?
$${\log }_2<!--\; \; -->(3f(n{)}^n)$$
P.S. This part of an assignment I'm working on, please do not give solutions

Can $(x+5{)}^{2n}+50=(x+20{)}^n+100$ be simplified algebraically to give a solution?
I have a brainteaser book which I'm working through and going overboard on. One puzzle is as follows: Your boss offers you a choice of two bonuses (1) fifty dollars after six months and an ongoing semiannual increase of five dollars; or (2) one hundred dollars after a year and an ongoing annual increase of twenty dollars. Which bonus will prove more lucrative? Rather than writing out a table, I wanted to calculate at what time one or the other becomes more lucrative.
I wrote out:
$$(x+5{)}^{2n}+50=(x+20{)}^n+100$$
, where $n=$ number of years (assuming $n\ge <!--\; \ge \; -->1$)
Then tried taking the log of both sides, but end up with:
$$2n\log <!--\; \; -->(x+5)=\log <!--\; \; -->[(x+5{)}^n+50]$$
How would I proceed from here, or must it be solved empirically?
Thanks in advance.

The position of significant digits and Logarithms relationship.....
I am unable to solve the following question as i don't understand what the relationship is between significant figures and Logarithms.
Q-If ${\log }_{10}<!--\; \; -->(7)=0.8451$ then the position of the first significant figure of $7^{-<!--\; -\; -->20}$.
The answer is the position of the first significant figure is 17th.
My book solves it in the following method-
$${\log }_{10}<!--\; \; -->(x)=-<!--\; -\; -->20{\log }_{10}<!--\; \; -->(7)$$
$$=-<!--\; -\; -->16.9020=-<!--\; -\; -->17+1-<!--\; -\; -->0.9020=\stackrel{¯<!--\; ¯\; -->}{1}7.0980$$
so the position is 17.
I fail to understand how and why this has happened please explain this solution to me and the relationship between significant figures and Logarithms...

Natural logarithmic derivative trick
I was wondering if someone could just explain something.
If I have a function which is dependent on $x$, the familiar $f(x)$. Now, if I take the derivative of this, and multiple by $x$ and divide through by $f(x)$
How does this then become true:
$$\frac{x}{f(x)}\frac{d(f(x))}{dx}=\frac{d\ln <!--\; \; -->f(x)}{d\ln <!--\; \; -->x}$$
I'm thinking I'm either a.) tired, b.) stupud or c.) both

Is it correct to say that ${\log }_2<!--\; \; -->0=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$?
The logarithm is not defined at $x=0$, because it tends to $-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$ as x tends to 0 from above. But is it nevetheless correct to say that
$${\log }_2<!--\; \; -->(0)=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}?$$
Or is it better to say/ write
$${\log }_2<!--\; \; -->(0)=\underset{x\downarrow <!--\; \downarrow \; -->0}{lim}{\log }_2<!--\; \; -->(x)=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$$

Violin String PDE Modeling
If you pluck a violin string, and then finger the string, fixing it precisely in the middle, the tone increases by one octave. In mathematical terms this means that the frequency is doubled. Explain why this happens.
Plucking the string without fingering should be represented as
$$u_{tt}=c^2{u}_{xx},u(0,t)=0,u(L,t)=0,u(x,0)=f(x),u_t(x,0)=0$$
Plucking the string and then fingering should be represented as
$$v_{tt}=c^2{v}_{xx},v(0,t)=0,v(\frac{L}{2},t)=0fort\ge <!--\; \ge \; -->t_1,v(L,t)=0,v(x,0)=f(x),v_t(x,0)=0$$
My game plan was to first solve for u, then find the first time at which $u=0$ at $\frac{L}{2}$ to use as $t_1$, and finally solve for v after $t_1$ is known. This means that the string is plucked from position f(x) and allowed to follow its natural trajectory only until the first time the string crosses the x-axis at $\frac{L}{2}$, and is then pinned on the x-axis forever. However, the PDE in v seems to have too many conditions for its order. Furthermore, the conditions of the PDE in v are a superset of the conditions of the PDE in u, so I think the solutions to the PDE in v should be a subset of the solutions to the PDE in u, even if it is somehow possible to apply this many conditions to the PDE in v. Are these PDEs possible to solve? Am I on the right track with this approach to the exercise?

Find the value of k so that (-1, 2), (2, 3) and (4, k) are collinear.

Given $f(x)=2x-5$ and $g(x)=4x-1$, evaluate the composite function.
$f[g(x)]$