Expert Answers to Algebra 2 Problems

What number gives a result of −5 when 10 is subtracted from the quotient of the number and 5 ?

Modeling differential equation
I am trying to model some differential equation which can explain buying some assets that can increase itself like a virus.
So, let's assume my money is y(t), and number of assets is n(t).
- y(t) to a certain level c, it will increase the n(t) by +1
- n(t) generate profit r %
so I can see something like
$$\begin{gathered} dy(t)/dt=y_0+r\cdot <!--\; \cdot \; -->n(t) \\ dn(t)/dt=If(y(t)\ge <!--\; \ge \; -->c,y(t)-<!--\; -\; -->20andn+1,n) \end{gathered}$$
However, I have no clue how I can model this discrete-continuous variable differential equation. Is there any way to model the system and get the solution?

Modeling with equations riddle
A father said that sevens years ago, he was eleven times as old as his daughter. Now he is four times as old as she is. How old is the father?
Can this be solved as a system of equations? I am stuck with this problem because I don't think that I am properly setting up the model.
$FathersAge-<!--\; -\; -->7=11daughtersage$
$4daughtersage=Fathersage$

Is it meaningful to calculate $(x+1{)}^{x+2}$ in $GF(3^2)$, e.g. using discrete logs?
I constructed GF$3^2$ using the polynomials degree $<2$ and arithmetic modulo the irreducible polynomial $(x^2+1)$. Then I built a table of logs to the base of the primitive element $(2x+2)$.
I can see that the log function maps only the non-zero elements of the field, so it seems sensible to use modulo 8 arithmetic between log values. This seems to work - for example:
$\begin{array}{rl}{\log }_{2x+2}<!--\; \; -->({(2x+1{)}^2}_{mod{x}^2+1})& =(2\times <!--\; \times \; -->{\log }_{2x+2}<!--\; \; -->(2x+1){)}_{mod8}\\ {\log }_{2x+2}<!--\; \; -->(x)& =(2\times <!--\; \times \; -->(7){)}_{mod8}\\ 6& =6\end{array}$
But how should I calculate, for example, $(x+1{)}^{(x+2)}$? Clearly modulo 8 arithmetic isn't going to do the trick. I guess that I need to use an irreducible polynomial but I am uncertain as to which one. Can I just use $(x^2+1)$ again?

Is there a relation between logical consequence, validity (tautology), and formula modeling?
I am new to propositional logic and I have noticed that these 3 things use the same symbol (double turnstile) and I am wondering if they are related to each other somehow.
to say that a formula B is a logical consequence of another formula A we write: $A\models B$.
also if we want to say that a valuation V models the formula A we write: $V\models A$.
and if we want to say that the formula A is valid we write: $\models A$.
so, here are my questions:
1. is there a reason they all use the same symbol?
2. can we define validity and formula modeling in terms of the logical consequence?
3. can a formula be a logical consequence of a valuation?

Find the principal P that will generate the given future value A, where $A=\$11000$ at 9% compounded daily for 10 years.

Finding the limit of $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{n^{log(n)}}{(\log <!--\; \; -->n{)}^n}$
I try to calculate the following limit:
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{n^{\log <!--\; \; -->(n)}}{(\log <!--\; \; -->n{)}^n}$
I tried this:
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{e}^{(\log <!--\; \; -->(n){)}^2-<!--\; -\; -->n\log <!--\; \; -->(\log <!--\; \; -->(n))}$
Is this useful? & what do now?

${\ln }^2<!--\; \; -->(x)\stackrel{?}{=}2\ln <!--\; \; -->(x)$
Is it the same as $\ln <!--\; \; -->(x{)}^2$? And if so, is it equal to $2\ln <!--\; \; -->(x)$?
Thanks in advance.

Value of $\frac{1}{{\log }_a<!--\; \; -->abc}+\frac{1}{{\log }_b<!--\; \; -->abc}+\frac{1}{{\log }_c<!--\; \; -->abc}$
I guess the answer will be 1. But I don't know how to evaluate it. Can someone give me some tips?

Question about inequality and logarithms
Let's say we have an inequality:
$x<y$
does this hold true?
$\log <!--\; \; -->x<\log <!--\; \; -->y$
and if so, how can I prove it?

Calculate Ln$(i^i)$
My attempt:
Ln$(z)$=$\ln <!--\; \; -->|z|+i\arg <!--\; \; -->z$
$z=0+i^i=0+i\cdot <!--\; \cdot \; -->i$
$$\begin{gathered} |z|=\sqrt{0^2+i^2}=i \\ \arg <!--\; \; -->z=\arctan <!--\; \; -->(i/0) \end{gathered}$$
$1.$ how it can be that the modulus equal to $i$?
$2.$ how can I find the argument?

Simplifying Logs
Simplify:
$\frac{\log <!--\; \; -->a+\log <!--\; \; -->b-<!--\; -\; -->\log <!--\; \; -->c}{\log <!--\; \; -->d^2}$
Using the basic properties of logs, the numerator should simplify to $\log <!--\; \; -->(ab/c)$, if I'm not mistaken. The denominator $\log <!--\; \; -->d^2=2\log <!--\; \; -->d$ but I don't know where to go from there. Can it be further simplified?

How to prove this logarithm equation?
Given :
${\log }_{12}<!--\; \; -->18=a\text{ and }{\log }_{24}<!--\; \; -->54=b$
prove that:
$ab+5(a-<!--\; -\; -->b)=1$
My attempt: I couldn't solve it in any way, as base were not common. I could solve it if base of second equation was $12$ raised to something, but here it is $12\times <!--\; \times \; -->2$ :( Any help will be greatly appreciated.

A company manufactures and sells x transistor radios per week. The weekly cost and price demand equations are $C(x)=5000+2x$ and $p=10-<!--\; -\; -->0.001x$ where x can be no more than 10000 units. The government has decided to tax the company $2 for each radio produced. Taking into account this additional cost, how many radio should the company manufacture each week in order to maximize its weekly profit? What is the maximum weekly profit? How much should the company charge for the radios to realize the maximum weekly profit?

What is the remainder when $1^n+2^n+3^n+\dots <!--\; \dots \; -->+{99}^n$ is divided by $1+2+3+\dots <!--\; \dots \; -->+99$? My first idea on how to approach was to create a polynomial expansion for the first few terms and then try to find a pattern for the rest but this became cumbersome and I don't think that this is a right approach as this would quickly become a problem that involves factorials and I have not covered modular arithmetic, is there a way to approach this problem in such a way that does not involve using factorials with mods.

Is there a simple algorithm for exponentiating large numbers to large powers?
I've been thinking about this for some days, a multiplication is a lot of sums, so:
$75\times <!--\; \times \; -->75=\stackrel{\text{75 times}}{\stackrel{⏞<!--\; ⏞\; -->}{75+75+75+75+75+75+75+75+\cdots <!--\; \cdots \; -->}}$
But then, there is a simple algorithm that enable us to multiply without having to sum all those numbers. In the same way, exponentiation is repeated multiplication - but I wasn't taugth about such algorithm for exponentiation. I've been thinking in representing the number with a polynomial, for example:
$1038={10}^3+3\cdot <!--\; \cdot \; -->{10}^1+8\cdot <!--\; \cdot \; -->{10}^0$
Then:
${1038}^{1038}=({10}^3+3\cdot <!--\; \cdot \; -->{10}^1+8\cdot <!--\; \cdot \; -->{10}^0{)}^{1038}$
But from here (considering what I currently know) I'd have to multiply it $1038$ times. The mentioned multiplication would be:
$({10}^3+3\cdot <!--\; \cdot \; -->{10}^1+8\cdot <!--\; \cdot \; -->{10}^0{)}^{1038}=\stackrel{\text{1038 times}}{\stackrel{⏞<!--\; ⏞\; -->}{({10}^3+3\cdot <!--\; \cdot \; -->{10}^1+8\cdot <!--\; \cdot \; -->{10}^0)\cdot <!--\; \cdot \; -->({10}^3+3\cdot <!--\; \cdot \; -->{10}^1+8\cdot <!--\; \cdot \; -->{10}^0)\cdot <!--\; \cdot \; -->\dots <!--\; \dots \; -->}}$
The first idea I had: There might be some connection with the binomial theorem, but I don't see how it fits. The second idea would be to find a way to write:
$({10}^3+3\cdot <!--\; \cdot \; -->{10}^1+8\cdot <!--\; \cdot \; -->{10}^0{)}^{1038}$
In which the red exponents are multiplied in some way by $1038$. There might be some connection with logarithms here, but I don't see it. And it could be the case that these techniques won't yield the results I'm looking for, so: Is there a simple algorithm for large numbers elevated to large exponents?

Could somebody validate my proof regarding the limit of $\ln <!--\; \; -->(x_n)$ when, $x_n$ tends to $a$?
So, let me cearly state the problem:
Let $(x_n)$ be a convergent sequence, with: $x_n>0$, $\mathrm{\forall <!--\; \forall \; -->}n$, n natural number, and $x_n\to <!--\; \to \; -->a$, with $a>0$. Then $\ln <!--\; \; -->x_n\to <!--\; \to \; -->\ln <!--\; \; -->a$
Here is my idea for a proof:
Our goal is to proof that there $\mathrm{\forall <!--\; \forall \; -->}\mathrm{ϵ<!--\; ϵ\; -->}>0$ there is some $n_{\mathrm{ϵ<!--\; ϵ\; -->}}$, such that $\mathrm{\forall <!--\; \forall \; -->}n\ge <!--\; \ge \; -->n_{\mathrm{ϵ<!--\; ϵ\; -->}}$, we have that $|\ln <!--\; \; -->x_n-<!--\; -\; -->\ln <!--\; \; -->a|<\mathrm{ϵ<!--\; ϵ\; -->}$
So here is what I did. First:
$|\ln <!--\; \; -->x_n-<!--\; -\; -->\ln <!--\; \; -->a|=|\ln <!--\; \; -->\frac{x_n}{a}|$
Then, beacause $\ln <!--\; \; -->x<x$, $\mathrm{\forall <!--\; \forall \; -->}x>0$, it easily follows that $|\ln <!--\; \; -->x|<x$, $\mathrm{\forall <!--\; \forall \; -->}x>0$. Applying this, we have that:
$|\ln <!--\; \; -->x_n-<!--\; -\; -->\ln <!--\; \; -->a|=|\ln <!--\; \; -->\frac{x_n}{a}|<\frac{x_n}{a}=\frac{x_n-<!--\; -\; -->a+a}{a}=\frac{x_n-<!--\; -\; -->a}{a}+1$
Now, we use the basic property of the absolute value: $x_n-<!--\; -\; -->a\le <!--\; \le \; -->|x_n-<!--\; -\; -->a|$, that gives us:
$|\ln <!--\; \; -->x_n-<!--\; -\; -->\ln <!--\; \; -->a|<\frac{x_n-<!--\; -\; -->a}{a}+1\le <!--\; \le \; -->\frac{|x_n-<!--\; -\; -->a|}{a}+1$
Now, we use the fact that $x_n\to <!--\; \to \; -->a$. So, $\mathrm{\forall <!--\; \forall \; -->}\mathrm{ϵ<!--\; ϵ\; -->}>0$ there is some $n_{\mathrm{ϵ<!--\; ϵ\; -->}}$, such that $\mathrm{\forall <!--\; \forall \; -->}n\ge <!--\; \ge \; -->n_{\mathrm{ϵ<!--\; ϵ\; -->}}$, we have that $|x_n-<!--\; -\; -->a|<\mathrm{ϵ<!--\; ϵ\; -->}$. We choose an epsilon that takes the form $a({\mathrm{ϵ<!--\; ϵ\; -->}}_0-<!--\; -\; -->1)$. This choice is possible for any ${\mathrm{ϵ<!--\; ϵ\; -->}}_0$
Now, we have managed to obtain that:
$|\ln <!--\; \; -->x_n-<!--\; -\; -->\ln <!--\; \; -->a|<\frac{a({\mathrm{ϵ<!--\; ϵ\; -->}}_0-<!--\; -\; -->1)}{a}+1={\mathrm{ϵ<!--\; ϵ\; -->}}_0,\mathrm{\forall <!--\; \forall \; -->}n\ge <!--\; \ge \; -->n_{\mathrm{ϵ<!--\; ϵ\; -->}}$
Since $\mathrm{\forall <!--\; \forall \; -->}{\mathrm{ϵ<!--\; ϵ\; -->}}_0>0$ we can find an $n_{\mathrm{ϵ<!--\; ϵ\; -->}}$, such that $\mathrm{\forall <!--\; \forall \; -->}n\ge <!--\; \ge \; -->n_{\mathrm{ϵ<!--\; ϵ\; -->}}$, the above inequality is staisfied, our claim is proved.
So, can you please tell me if my proof si correct? I've tried to find a proof, only for the limit of sequences! This problem has been on my nerves for s while. Also, probably there is some simpler way to do it, but I couldn't find it.

Rewriting in $y=A_0\cdot <!--\; \cdot \; -->e^{at}$
How do you rewrite $y=-8(1.589{)}^{t-3}$ in $y=A_0{e}^{at}$ form for appropriate constants $A_0$ and $a?$
For other problems I took the $\ln$ of the number inside the parenthesis. So for example I would've done
$y=-<!--\; -\; -->8\ln <!--\; \; -->(1.589{)}^{t-<!--\; -\; -->3}$
then I would get the answer
$y=-<!--\; -\; -->8(e^{0.46317-<!--\; -\; -->3})$
but that is wrong on my assignment. What do I do with the $-<!--\; -\; -->3?$ I tried multiplying it by the $-<!--\; -\; -->8$ to get $24$ but that isn't right either.

How to compute the derivative of ${\sqrt{x}}^{\sqrt{x}}$?
I know have the final answer and know I need to use the natural log but I'm confused about why that is.
Could someone walk through it step by step?

Dividing logarithms without using a calculator
The problem I have is: $\frac{\log <!--\; \; -->16+\log <!--\; \; -->25-<!--\; -\; -->\log <!--\; \; -->36}{\log <!--\; \; -->10-<!--\; -\; -->\log <!--\; \; -->3}$
(log is base 10 here)
I have the answer as 2 but no idea how to reach it..
I need to work this out without the use of a calculator but I can't get my head round it. I know that
$\log <!--\; \; -->(16)=\log <!--\; \; -->(4^2)=2(\log <!--\; \; -->4)$ $\log <!--\; \; -->25=2(\log <!--\; \; -->5)$
I can add \logs together when they all are a power of the same number e.g.
$\log <!--\; \; -->(64)+\log <!--\; \; -->(32)=\log <!--\; \; -->(2^6)+\log <!--\; \; -->(2^5)=6(\log <!--\; \; -->2)+5(\log <!--\; \; -->2)=11(\log <!--\; \; -->2)$
I might just be over thinking it or over complicating it but I'd really appreciate some help here. Thanks.