Expert Answers to Algebra 2 Problems

Get d from Free space path loss equation
How would I get $d$ from this: $L=10n{\log }_{10}<!--\; \; -->(d)+C$
I've got this so far:
$\begin{array}{rl}L-<!--\; -\; -->C& =10n{\log }_{10}<!--\; \; -->(d)\\ \frac{L-<!--\; -\; -->C}{10n}& ={\log }_{10}<!--\; \; -->(d)\end{array}$
Is it: $2^{10}\cdot <!--\; \cdot \; -->\frac{L-<!--\; -\; -->C}{10n}=d$???

How do I solve for x in $\ln <!--\; \; -->(x)\ln <!--\; \; -->(x)=2+\ln <!--\; \; -->(x)$

How is the Logarithm derived from the exponential function? (aren't they inverses?)
I've been learning logs in school, and my teacher, friend, and I are stumped on something. How does one derive the logarithmic function from the exponential function? My friend thinks Tayler Series are the trick. Is he right? Is there a a better/simpler/more elegant way? Also, do calculators use taylor series to do logs? Thanks for the help

What's the formula to solve summation of logarithms?
I'm studying summation. Everything I know so far is that:
$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}k=\frac{n(n+1)}{2}$
$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{k}^2=\frac{n(n+1)(2n+1)}{6}$
$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{k}^3=\frac{n^2(n+1{)}^2}{4}$
Unfortunately, I can't find neither on my book nor on the internet what the result of:
$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}\log <!--\; \; -->i$
$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}\ln <!--\; \; -->i$
is.
Can you help me out?

An integral involving the inverse of $f(x)=\log <!--\; \; -->x-<!--\; -\; -->\log <!--\; \; -->\cos <!--\; \; -->x+x\tan <!--\; \; -->x$
Let the function $f:(0,{\displaystyle \frac{\mathrm{\pi <!--\; \pi \; -->}}{2}})\to <!--\; \to \; -->\mathbb{R}$ be defined as
$f(x)=\log <!--\; \; -->x-<!--\; -\; -->\log <!--\; \; -->\cos <!--\; \; -->x+x\tan <!--\; \; -->x.$
Let its inverse be denoted as $f^{(-<!--\; -\; -->1)}:\mathbb{R}\to <!--\; \to \; -->(0,{\displaystyle \frac{\mathrm{\pi <!--\; \pi \; -->}}{2}})$, it satisfies
$\mathrm{\forall <!--\; \forall \; -->}z\in <!--\; \in \; -->\mathbb{R},f(f^{(-<!--\; -\; -->1)}(z))=z.$
Consider the integral
$I={\int <!--\; \int \; -->}_0^{1}z{e}^z{f}^{(-<!--\; -\; -->1)}(z)dz.$
Is it possible to evaluate the integral $I$ in a closed form?

Suppose $f\in <!--\; \in \; -->O_p(V)$ is a rational function on V that has a value at $p$. Then write $f=a/b=a^{\prime }/b^{\prime }$ where $a,b,a^{\prime },b^{\prime }\in <!--\; \in \; -->\mathrm{\Gamma <!--\; \Gamma \; -->}(V)$, the coordinate ring of V. Want to show the value of $f$ at $p$ is well-defined, i.e. $a(p)/b(p)=a^{\prime }(p)/b^{\prime }(p)$.
So since $a/b=a^{\prime }/b^{\prime }$ are the same equivalence class, there is some non-zero poly $x\in <!--\; \in \; -->\mathrm{\Gamma <!--\; \Gamma \; -->}(V)$ such that $x(a{b}^{\prime }-<!--\; -\; -->a^{\prime }b)=0$. Then $x(p)(a(p)b^{\prime }(p)-<!--\; -\; -->a^{\prime }(p)b(p))=0$. Then what? How do we know $x(p)\ne <!--\; \ne \; -->0$?

How solve this logarithms equation
What relationship between a,b and c ?
${\log }_a<!--\; \; -->u+{\log }_b<!--\; \; -->u+{\log }_c<!--\; \; -->u={\log }_{abc}<!--\; \; -->u$

I know that an analytic function on $\mathbb{C}$ with a nonessential singularity at $\mathrm{\infty <!--\; \infty \; -->}$ is necessarily a polynomial.
Now consider a meromorphic function $f$ on the extended complex plane $\stackrel{^<!--\; ^\; -->}{\mathbb{C}}$. I know that $f$ has only finitely many poles, say $z_1,\dots <!--\; \dots \; -->,z_n$ in $\mathbb{C}$. Suppose also that $f$ has a nonessential singularity at $\mathrm{\infty <!--\; \infty \; -->}$.
Then if $z_i$ have orders $n_i$, it follows that $\prod <!--\; \prod \; -->(z-<!--\; -\; -->z_i{)}^{n_i}f(z)$ is analytic on $\mathbb{C}$, and has a nonessential singularity at $\mathrm{\infty <!--\; \infty \; -->}$, and is thus a polynomial, so $f$ is a rational function.
But I'm curious, what if f doesn't have a singularity at $\mathrm{\infty <!--\; \infty \; -->}$, or in fact has an essential singularity at $\mathrm{\infty <!--\; \infty \; -->}$ instead? Is $f$ still a rational function?

How to solve the following equation?
I have been given this equation for homework and I don't know how to solve it.
$y={\displaystyle \frac{\ln <!--\; \; -->({\displaystyle \frac{x}{m}}-<!--\; -\; -->sa)}{r^2}}$

What is the difference between logarithmic decay vs exponential decay?
I am a little unclear on whether they are distinctly different or whether this is a 'square is a rectangle, but rectangle is not necessarily a square' type of relationship.

I was looking into a previous exam from 2011 of a course I am taking of Complex Analysis, and they ask
Which of the following series converge to a rational function in some domain?
$$\begin{gathered} \underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{k!+k^2+k}{z}^{k^2+2k} \\ \underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{2^k}{(1+z^2{)}^k} \\ \underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{k!(z-<!--\; -\; -->k{)}^k} \end{gathered}$$
I had no problem with the second and third ones. The second one converges to $\frac{1}{1-<!--\; -\; -->\frac{2}{1+z^2}}$ given that $\left|\frac{2}{1+z^2}\right|<1$, and the third one has an infinite number of singularities so it can't be a rational function.
I don't know how to verify that in the first series. Is there any general method to see if some power series converge to a rational function, or in this particular case, a way to see if this series does?
Edit: Now I am not that convinced about my argument for the third series (Since the series might only converge in a bounded domain from which the number of singularities would be finite). Is there anything wrong with it or any way to formalize it further?

I have a small question:
Can one state that
$f(x)=\frac{1}{x}$
is a rational function because it is the quotient between a polynomial with degree 0 and a polynomial with degree 1?
Thanks!

How does $7\log <!--\; \; -->(8x)=7\ln <!--\; \; -->8x$
I was working on some math homework with a program called scientific notebook. I was check that
I was writing something correctly.
The original equation is $(\log <!--\; \; -->(x^4)+\log <!--\; \; -->(x^5))/\log <!--\; \; -->(8x)=7$
I then converted it to $\log <!--\; \; -->(x^{(4+5)})=7\log <!--\; \; -->(8x)$
I was expecting to get $\log <!--\; \; -->((8x{)}^7)$ when I entered the $7\log <!--\; \; -->(8x)$ into the program to evaluate. Can someone please explain this to me?

It's a logarithmic worksheet and O can't solve it.
${\log }_a<!--\; \; -->x=p$, ${\log }_b<!--\; \; -->x=q$ , ${\log }_{abc}<!--\; \; -->x$=r. What is ${\log }_c<!--\; \; -->x$?.. It's on my math homework can someone solve it cause I need it.

Using Logarithms
$\begin{array}{rl}-<!--\; -\; -->2^{n-<!--\; -\; -->1}\ln <!--\; \; -->2& =-<!--\; -\; -->100\ln <!--\; \; -->10\\ & \\ -<!--\; -\; -->100\ln <!--\; \; -->10& =-<!--\; -\; -->230\\ & \\ {\displaystyle \frac{-<!--\; -\; -->230}{\ln <!--\; \; -->(2)}}& =-<!--\; -\; -->333\\ & \\ -<!--\; -\; -->2^{n-<!--\; -\; -->1}& >-<!--\; -\; -->333\\ & \\ (n-<!--\; -\; -->1)\ln <!--\; \; -->(-<!--\; -\; -->2)& >\ln <!--\; \; -->(-<!--\; -\; -->333)\end{array}$
Here is where I am stuck.
I am not sure if this part is correct: $-<!--\; -\; -->n-<!--\; -\; -->1=8$. Then solving we would get $-<!--\; -\; -->9$

Let $f(z)$ be a rational function in the complex plane such that $f$ does not have any poles in $\{z:\mathrm{\Im <!--\; \Im \; -->}z\ge <!--\; \ge \; -->0\}$.
Prove that $sup\{|f(z)|:\mathrm{\Im <!--\; \Im \; -->}z\ge <!--\; \ge \; -->0\}=sup\{|f(z)|:\mathrm{\Im <!--\; \Im \; -->}z=0\}$.
Let ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_r$ be a half circle counter such that ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_r={\mathrm{\Gamma <!--\; \Gamma \; -->}}_{r_1}+{\mathrm{\Gamma <!--\; \Gamma \; -->}}_{r_2}$ when ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_{r_1}=\{z:\mathrm{\Im <!--\; \Im \; -->}z=0,|z|=r\}$, ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_{r_2}=\{z:\mathrm{\Im <!--\; \Im \; -->}z>0,|z|=r\}$. Using the Maximum modulus principle on the insides of ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_r$, $|f|$ Gets is maximum value on ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_r$. As r gets bigger if |f| got it's maximum on ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_{r_2}$ than it's still smaller the the value on ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_{r+1}$ which does not contain ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_{r_2}$. I would like a hint on how to proceed.

What calculate $\ln <!--\; \; -->i$
I would like to know how to calculate $\ln <!--\; \; -->i$. I found a formula on the internet that says
$\ln <!--\; \; -->z=\ln <!--\; \; -->|z|+i\mathrm{Arg}<!--\; \; -->(z)$
Then $|i|=1$ and $\mathrm{Arg}<!--\; \; -->(i)$ is?

logarithms equations, different bases
solve equations: ${\log }_x<!--\; \; -->10+2{\log }_{10x}<!--\; \; -->10-<!--\; -\; -->3{\log }_{100x}<!--\; \; -->10=0$ so I tried to use ${\log }_a<!--\; \; -->b=\frac{1}{{\log }_b<!--\; \; -->a}$ but it didn't work for me.

Am I correctly determining the convergence or divergence of these series:
$\sum <!--\; \sum \; -->\frac{1}{n(\ln <!--\; \; -->(n){)}^{5/4})}$