Expert Answers to Algebra 2 Problems

You need 60 mL of a 45% alcohol solution. On hand, you have a 30% alcohol mixture. You also have a 50% alcohol mixture. How much of each mixture will you need to add to obtain the desired solution?

Help with radical logs. ${\log }_2<!--\; \; -->\sqrt{32}$
I am having trouble understanding logarithms. Specifically I can't understand this equation.
$${\log }_2<!--\; \; -->\sqrt{32}$$
I know the answer is $\left(\frac{5}{2}\right)$ but I don't know how it is the answer. If somebody could please explain the process step by step for solving this log and perhaps give examples of other radical logs I would really appreciate it.

How to find all the intersection points of the two functions $\log <!--\; \; -->(x!)$ and x?
I am trying to find where $\log <!--\; \; -->(x!)$ and x intersect, and am unable to do so rigorously.
I eventually have $2^x=x!$, but I am unsure how to proceed from here.
Any input as to how to go about solving this would be great.

A pilot can travel 450miles with the wind in the same amount of time as 360 miles against the wind. Find the speed of the wind if the​ pilot's speed in still air is 315miles per hour.

(a) What is a system of equations in the variables x, y, and z? (b) What are the three methods we use to solve a system of equations?

Find bif one of the solutions of the equation $$x^2+bx+1=0$$ twice the other.

What will be the value of this integral
$${\int <!--\; \int \; -->}_0^{\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}}\left[e^{\ln <!--\; \; -->(\cos <!--\; \; -->x\cdot <!--\; \cdot \; -->\frac{d(\cos <!--\; \; -->x)}{dx})}\right]dx$$
We know that $a^{lo{g}_a(c)}=c$.
But in this question, the expression in logarithm function evaluates to be negative, for which log is not defined.
$${\int <!--\; \int \; -->}_0^{\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}}\left[e^{\ln <!--\; \; -->(-<!--\; -\; -->\frac{sin(2x)}{2})}\right]dx$$
So this means we cannot apply the log property? my working:
$$={\int <!--\; \int \; -->}_0^{\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}}[\cos <!--\; \; -->x\cdot <!--\; \cdot \; -->(-<!--\; -\; -->\sin <!--\; \; -->x)]dx$$
$$={\left[{\displaystyle \frac{{\cos }^2<!--\; \; -->x}{2}}\right]}_0^{\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}}$$
$$=\boxed{{\displaystyle -<!--\; -\; -->{\displaystyle \frac{1}{2}}}}$$
Is this integral, with undefined log, correct?

Complex but Simple equation.
Im( e^z ) = 1
I've reached rsin(t) = 1 but I am stuck here, where r is radius and t is argument.
There may be use of logs here I might be missing.

Solve the equation for x.
$$\ln <!--\; \; -->(\ln <!--\; \; -->x)=5$$

Solving logarithmic equation for $n$
$$8n^2=64n{\log }_2<!--\; \; -->n$$
Haven't done algebra in years and can't figure out how to get rid of the ${\log }_2$.

Logarithmic question
In the following question I fail to understand why the A option is correct.
I understand that D is wrong, and that B and C are correct, but why is A correct?
If $3^x=4^{x-<!--\; -\; -->1}$, then xcannot be equal to:
A. $2{\log }_3<!--\; \; -->2$
B. $\frac{2}{2-<!--\; -\; -->{\log }_2<!--\; \; -->3}$
C. $\frac{1}{1-<!--\; -\; -->lo{g}_{4}3}$
D. $\frac{2{\log }_2<!--\; \; -->3}{2{\log }_2<!--\; \; -->3-<!--\; -\; -->1}$
By taking ${\log }_3$ on both sides, we get $x=2(x-<!--\; -\; -->1){\log }_3<!--\; \; -->2$
After that we get B and C as correct after making modifications but how do we know that A is correct?

Danielle's text messaging plan charges 30¢ for each message over 350 in addition to a $13 base charge. If she owes $83.50 for text messaging in the month of April, how many text messages did she send that month?

VP Definition: Why polynomial bound on the degree?
The complexity class VP (Valiant P) is defined to be the class of all polynomials of polynomially bounded degree which can be realized by an arithmetic circuit family with polynomially bounded size.
My questions are:
1. Why do we require the degree to be polynomially bounded?
2. Which complexity class to we get if we omit this constraint?

Let P and Q be polynomials with integer coefficients such that P(a) divides Q(a) in Z for every integer a. Does it mean that P divides Q in Z[X]?
P(a) divides Q(a) means that there exists an integer k(a) such that $Q(a)=k(a)P(a)$, so $Q=kP$. This means that P divides Q if k is polynomial with integer coefficients. But I can't find a proof or a counter-example. Any idea?
For the context, I was looking at a proof of the fact that $X^2-<!--\; -\; -->X+1$ divides $X^{2n+1}+(X=1{)}^{n+2}$ for every $n\ge <!--\; \ge \; -->1$ (in a video of Michael Penn I think but I am not sure). The method used was to consider congruences modulo $P(X)=X^2-<!--\; -\; -->X+1$. This is fine and clever method, but my students don't know arithmetic in Z[X], so I tried to circumvent the problem by looking at congruence in Z modulo P(a) for every $a\in <!--\; \in \; -->\mathbb{Z}$. But I hit the problem stated above to conclude the reasoning.

Mathematical modeling and operational research
I have a variable TE which $0\le <!--\; \le \; -->TE\le <!--\; \le \; -->180$
Also, I have a parameters L(k) and U(K), showing lower bound and upper bound for interval K $(k=1,2,3)$
The value of these parameters are as follows: $L(1)=0K(1)=60,L(2)=60K(2)=120,L(3)=120K(3)=180$
I need to find the related range of the variable TE. For example,
if `$TE=12$`, then `$K=1$`
if `$TE=150$`, then `$k=3$`
Which mathematical constraint(s) do this?

Is the Polynomial $X^{p^n}-<!--\; -\; -->X$ the zero polynomial in characteristic p?
Suppose that p is a prime number and that ${\mathbb{F}}_{p^n}[X]$ is the ring of polynomials with one variable, with coefficients in the field ${\mathbb{F}}_{p^n}$ for some $n\in <!--\; \in \; -->\mathbb{N}$. Then
$p(X):=X^{p^n}-<!--\; -\; -->X\in <!--\; \in \; -->{\mathbb{F}}_{p^n}[X]$
is a polynomial, which appears at many places in number theory, for example in the definition of the Carlitz Exponential. (In books about function field arithmetics this polynomial is often called [1], where in general $[k]:=X^{p^{n\cdot <!--\; \cdot \; -->k}}-<!--\; -\; -->X$)
Now the question is: Since $p(X)=0$ for any $X\in <!--\; \in \; -->{\mathbb{F}}_{p^n}$, isn't f just the zero polynomial? The same should be true for all [k], but as this is not mentioned in any book on function field arithmetic, I think I might overlook something.
Maybe someone has something enlightening to say here.

Exponential function leads to unsolvable logarithm.
Problem:$f(t)=0.75\ast <!--\; \ast \; -->{10}^{kt}$ where k is a constant. We know that $f(2)=3$. Find $f(3)$
My approach:
Rewrite $f(2)$ as: $0.75\ast <!--\; \ast \; -->{10}^{2k}=3$
Try to solve for k, eliminate 0.75 first: ${10}^{2k}=4$
This is where I get stuck: $lo{g}_{10}4=2k$
Where do I go from there? Am I doing something wrong? The answer is supposed to be $f(3)=6$
Edit: typo in the logs, fixed. And I solved it, thanks everyone!

The voltage v(t) in a certain circuit is given by the equation:
$$v(t)=4t^3-<!--\; -\; -->16t^2+19t-<!--\; -\; -->6\text{ for }0\le <!--\; \le \; -->t\le <!--\; \le \; -->3$$ seconds
Determine the global maximum and the global minimum for v(t) on the time interval $$0\le <!--\; \le \; -->t\le <!--\; \le \; -->3$$ seconds.

The length of a rectangle is 5 in. more than twice its width. Its area is 102 in^2. Find the width of the rectangle.

Determine the convergence or divergence of $\underset{2}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{(\log <!--\; \; -->n{)}^s}$, where $s\in <!--\; \in \; -->\mathbb{R}$ is given.
Since
$$\frac{1}{(\log <!--\; \; -->n{)}^s}>\frac{1}{n^s}$$
for large $n$, if $s\le <!--\; \le \; -->1$ then $\underset{2}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{(\log <!--\; \; -->n{)}^s}$ diverges.
But for $s>1$ I have not yet figured out a proof.