Expert Answers to Algebra 2 Problems

Determining the value of parameters given constraints
If
$$\frac{x(y+z-<!--\; -\; -->x)}{\log <!--\; \; -->x}=\frac{y(z+x-<!--\; -\; -->y)}{\log <!--\; \; -->y}=\frac{z(x+y-<!--\; -\; -->z)}{\log <!--\; \; -->z}$$
and
$$a{x}^y{y}^x=b{y}^z{z}^y=c{z}^x{z}^y$$
then what is the value of $a+\frac{b}{c}$?
I am getting as $a{x}^y{y}^x=b{y}^z{z}^y=c{z}^x{x}^z$ after solving the first equation but according to the question it is not correct. So now I have to proove only that $x^z=z^y$, then my answer will be equal to $2$.

How to prove that $f(x)=x^{\epsilon }-<!--\; -\; -->\log <!--\; \; -->x$ is $\mathrm{\infty <!--\; \infty \; -->}$ when $x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$?
I'm trying to prove that the function $x^{\epsilon }$ is "bigger" than $\log <!--\; \; -->x$ when $x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$, for every $\epsilon >0$
Or to put it in a more formal way:
For every $\epsilon >0$, there exists a constant $N$ for which all $x>N$ the following is true:
$x^{\epsilon }>\log <!--\; \; -->x$
I don't believe that the base of the log is important for this one, but just in case, it is 2.

In the spring of 2010 an off-shore oil drilling rig exploded in the Gulf of Mexico. Not long after the explosion there was a 2 mm thick oil slick that was 8.05 km long by 4.83 km wide. How much oil was in the slick? Express your answer in cubic meters

A system of linear equations in two variables may have infinitely many solutions. In such a case, the equations are said to be _________.

Simplify and do NOT leave negative exponents in your answer.
$$(\frac{4x^7{y}^8}{-<!--\; -\; -->16x^{12}{y}^{-<!--\; -\; -->3}}{)}^{-<!--\; -\; -->3}$$

Is there proof show that $\log <!--\; \; -->x$ is undefined and make no sense at $x=0$?
Is there proof show that $\log <!--\; \; -->x$ is undefined at $x=0$?
Note(01):: log is the inverse function of the exponential function.
note(02): I edited my question as I meant why it's not make a sens at $x=0$?
Thank you for your help .

All real number solutions of equation ${\log }_{2011}<!--\; \; -->(2010x)={\log }_{2010}<!--\; \; -->(2011x)$ are in certain interval. Which one is it?
This task has to be done with no calculator, but I don't have basic idea how to start. Can someone give me advice, I know this is pretty easy but I need direction for particularly this one? EDIT: I do it like this
$${\log }_{2011}<!--\; \; -->(2010x)={\log }_{2010}<!--\; \; -->(2011x)$$
$$\frac{\ln <!--\; \; -->2010x}{\ln <!--\; \; -->2011}=\frac{\ln <!--\; \; -->2011x}{\ln <!--\; \; -->2010}$$
$$\frac{\ln <!--\; \; -->2010+\ln <!--\; \; -->x}{\ln <!--\; \; -->2011}=\frac{\ln <!--\; \; -->2011+\ln <!--\; \; -->x}{\ln <!--\; \; -->2010}$$
$${\ln }^2<!--\; \; -->2010+\ln <!--\; \; -->x(\ln <!--\; \; -->2010)={\ln }^2<!--\; \; -->2011+\ln <!--\; \; -->x(\ln <!--\; \; -->2011)$$
$${\ln }^2<!--\; \; -->2010-<!--\; -\; -->{\ln }^2<!--\; \; -->2011-<!--\; -\; -->\ln <!--\; \; -->x(\ln <!--\; \; -->2011-<!--\; -\; -->\ln <!--\; \; -->2010)=0$$
$$(\ln <!--\; \; -->2011-<!--\; -\; -->\ln <!--\; \; -->2010)(\ln <!--\; \; -->2011+\ln <!--\; \; -->2010-<!--\; -\; -->\ln <!--\; \; -->x)=0$$
$$\ln <!--\; \; -->\frac{(2010\ast <!--\; \ast \; -->2011)}{x}=\ln <!--\; \; -->1$$
$$x=2010\ast <!--\; \ast \; -->2011$$
What am I doing wrong here? And also, how to even get solution (which is $x\in <!--\; \in \; -->(0,\frac{1}{2011}])$

Solution:
$$\sqrt[5]{x^3{y}^2}\sqrt[10]{x^4{y}^{16}}$$

Prove that $m-<!--\; -\; -->n$ divides p
Let $a_1,a_2,a_3$ be a non constant arithmetic Progression of integers with common difference p and $b_1,b_2,b_3$ be a geometric Progression with common ratio r. Consider 3 polynomials $P_1(x)=x^2+a_{1}x+b_1,P_2(x)=x^2+a_{2}x+b_2,P_3(x)=x^2+a_{3}x+b_3$. Suppose there exist integers m and n such that $gcd(m,n)=1$ and $P_1(m)=P_2(n),P_2(m)=P_3(n)$ and $P_3(m)=P_1(n)$. Prove that $m-<!--\; -\; -->n$ divides p.
I made the 3 equations but am unable to get the result from them. How should I manipulate them.

A school district has a student population of $36,734 students. If the maximum class size is 25 students. How many more teacher must the district have to supplement their staff of 217 teachers?

Solve for the unknown variable.
$$6e^{7x}=75$$

Simplifying a log of a log
I have a summation series that unfortunately involves a log of a log. It looks like the following (assume all $\log$ are log base $2$):
$$\underset{i=1}{\overset{k}{\sum <!--\; \sum \; -->}}\log <!--\; \; -->\log <!--\; \; -->\frac{n}{2^{k-<!--\; -\; -->i}}$$
I'd like to simplify this somehow, such as move the logs or the n value to the outside of the equation, so that I can use the closed form of the geometric series sum to further simplify the equation, but I'm not sure where to start.
Thanks for the help!

Write the expression as the sum or difference of two logarithmic functions containing no exponents.
$${\log }_9<!--\; \; -->(x\sqrt[3]{x+2})$$

Test whether a given function is polynomial
You have a black box function to which you can give real number inputs and from which you can receive real number outputs. How would you test whether it is likely to be a polynomial?
One expensive idea is to use finite differences:
1. Choose a maximum degree n of the "polynomial" you are testing.
2. Choose a consecutive sequence with random step size, and evaluate the function there to get an output sequence. E.g.,
$$[2,2.3,2.6,2.9,\dots <!--\; \dots \; -->]\to <!--\; \to \; -->[4.81,5.02,5.05,4.90,\dots <!--\; \dots \; -->]$$
3. Using the output sequence as S[0], define S[n] so that its k-th entry $S[n][k]=S[n-<!--\; -\; -->1][k+1]-<!--\; -\; -->S[n-<!--\; -\; -->1][k].E.g.S[1]=[5.02-<!--\; -\; -->4.81,5.05-<!--\; -\; -->5.02,4.90-<!--\; -\; -->5.05,...]=[0.21,0.03,-<!--\; -\; -->0.15,...]$
4. If the function is a polynomial (of degree at most n), then the sequence $S[n+1]$ should be all zeros.
Some issues about programming this method:
Would be expensive for large n
If S[0] has large values, computer arithmetic will produce bad results for S[1] and beyond.

Approximating the cosine by Taylor polynomial
Let $f:=\cos <!--\; \; -->(x)$ I'm asked to find for which values of x we can be sure the 4th degree Taylor polynomial will give an error lesser than $\frac{1}{1000}$.
Now, $\cos <!--\; \; -->(x)=1-<!--\; -\; -->\frac{x^2}{2}+\frac{x^4}{24}+o(x^4)$
Taylor's 4th degree remainder is then $R=\frac{f^{(5)}(\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}))}{5!}{x}^5=\frac{-<!--\; -\; -->\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})}{5!}{x}^5$
Since I'm only asked about the error, I can get rid of the sign, and can do the following boundering:
$|R|\le <!--\; \le \; -->|\frac{1}{5!}{x}^5|\le <!--\; \le \; -->\frac{1}{1000}$
Since that is strictly increasing, its easy to solve
that last inequation, by doing $\frac{1}{120}{x}^5=\frac{1}{1000}⟶<!--\; ⟶\; -->x=\sqrt[5]{\frac{120}{1000}}\approx <!--\; \approx \; -->0.65$
Which ends up giving $|R|\le <!--\; \le \; -->\frac{1}{1000}\mathrm{\forall <!--\; \forall \; -->}x:|x|\le <!--\; \le \; -->\sqrt[5]{\frac{120}{1000}}(\approx <!--\; \approx \; -->0.65)$
Now, I thought this was an already correct solution, but to my surprise, I checked plotted both the function and its Taylor polynomial, and the error bound that produces below 0.65 is not only lesser than $\frac{1}{1000}$, but actually lesser than $\frac{1}{10000}$
e.g:
$$\cos <!--\; \; -->(0.65)=\mathrm{0.796187...}$$
$$taylor(cos(0.65))=\mathrm{0.796083...}$$
Which means it is overly precise, what is going on here, I'm asuming it must be some arithmetic mistake I made somewhere, but I can't find it and its driving me crazy.

Solving a logarithmic expression without a calculator
How do I find the value of this logarithmic expression without using a calculator? I'm trying to relearn algebra, but this problem has me scratching my head, and my Google tutorial searches are failing me.
$2^{{\log }_2<!--\; \; -->10}$

Modeling Question: What to make my decision variables
I was given the following problem, and I am having trouble determining what to make my decision variables to model it:
Your goal is to optimize your profit. Over a 10 week period the demands for your product are 85, 97, 120, 105, 84, 74, 116, 123, 104, 98. The item sells for 50 dollars. There is a 500 dollar shipping charge on each order and if you order 150 or fewer items, each item costs 30. If you order between 150 and 249, each item costs 27 dollars. If you order 250 or more items, then the cost is 24 dollars per item. Each item in inventory costs 5 dollars due to storage space, capital spent, etc. Each item with a shortage (demand exceeds inventory) costs 15 dollars due to a loss of customer satisfaction, these items are never sold. What is your ordering policy?
I've tried calling $x_{ij}$ the number of products bought in week i at quantity j, where i goes from 1 to 10 and j goes from 1 to 3. Then I need a binary decision variable. I said let $b_{ij}=1$ if $x_{ij}>0$ and 0 if it is 0. I have issues modeling the inventory and shortage costs, then. There has to be a better way to define my decision variables.

Show that $\ln <!--\; \; -->(x)\le <!--\; \le \; -->x-<!--\; -\; -->1$
I'm not really sure how to show this, it's obvious if we draw a graph of it but that won't suffice here. Could we somehow use the fact that $e^x$ is the inverse? I mean, if $e^{x-<!--\; -\; -->1}\ge <!--\; \ge \; -->x$ then would the statement be proved?

Problem with this challenging summation
I'm having some trouble finding the summation of this series.
I tried all I could, but in the end the denominator is creating problem.
$$\underset{r=0}{\overset{n}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^r(\genfrac{}{}{0ex}{}{n}{r})\frac{1+r\ln <!--\; \; -->10}{(1+\ln <!--\; \; -->{10}^n{)}^r}$$
The answer is zero but in my book no solution is given.
Any help would be appreciated.

Attendance was 400 members. 75% voted to increase dues. What number of members voted for increasing dues?
Could you explain an easy way to calculate this?