Learn How to Graph Logarithmic Equations

Why does the branch cut for log(1+z), cutting away the negative axis, start at -1?
I am really confused about this concept.
Is it because from -1 to 0, the inputs (1+z) are just viewed as inputs that aren't actually from the negative axis at all, and so log(1+z) is well-defined and continuous on (-1,0]? I don't see the single-valuedness on (-1,0], though. Shouldn't log(1+z) jump by a $2\mathrm{\pi <!--\; \pi \; -->}i$ term, on (-1,0]?
Any comments are welcome.
Thanks,

How to prove "Logarithms grow more slowly than polynomials"
"Logarithms grow more slowly than polynomials. That is, $\Theta (\lg <!--\; \; -->n)$ grows more slowly than $\frac{-<!--\; -\; -->b\pm <!--\; \pm \; -->\sqrt{b^2-<!--\; -\; -->4ac}}{2a}$ for any positive constant a."
if $y_1=x^{1/2}$) and $y_2=\log <!--\; \; -->(x)$
by comparing the graphs can say $y_1>y_2$
But, is there a way to prove this?

logarithmic Series
I'm aware that by properties of logarithm
$\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}\ln <!--\; \; -->(k)=\ln <!--\; \; -->(n!)$
My question is if
$\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{\ln }^2<!--\; \; -->(k)={\ln }^2<!--\; \; -->(n!)?$
Because when I am verifying the value where $n=5$, I get different result... maybe Im missing something... is there a formula that defines
$\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{\ln }^2<!--\; \; -->(k)\text{ ?}$
Here's my computation:
$\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{\ln }^2<!--\; \; -->(k)={\ln }^2<!--\; \; -->(n!)$
$\underset{k=1}{\overset{5}{\sum <!--\; \sum \; -->}}{\ln }^2<!--\; \; -->(k)={\ln }^2<!--\; \; -->(5!)$
${\ln }^2<!--\; \; -->(1)+{\ln }^2<!--\; \; -->(2)+{\ln }^2<!--\; \; -->(3)+{\ln }^2<!--\; \; -->(4)+{\ln }^2<!--\; \; -->(5)={\ln }^2<!--\; \; -->(120)$
$0+0.48045+1.20694+1.92181+2.59029=22.92007$
$6.199494$ is not equal to $22.92007$

I need help solving a logarithm equation.
these were my steps. Can someone tell me where have i goe wrong since the answer is $-<!--\; -\; -->\frac{1}{4}$
$\sqrt{\log <!--\; \; -->(\sqrt{10}a)}=\frac{1}{2}$
$\frac{1}{2}\log <!--\; \; -->(\sqrt{10}a)=\log <!--\; \; -->\sqrt{10}$
${\log <!--\; \; -->(\sqrt{10}a)}^{\frac{1}{2}}=\log <!--\; \; -->\sqrt{1}0$ because $({\log }_a<!--\; \; -->x^n=n{\log }_a<!--\; \; -->x)$
$(\sqrt{10}a{)}^{\frac{1}{2}}={10}^{\frac{1}{2}}$ (squaring)
${10}^{\frac{1}{2}}a=10$ division by ${10}^{\frac{1}{2}}$
$a={10}^{1-<!--\; -\; -->\frac{1}{2}}$
$a=\sqrt{10}$

How to compute $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\displaystyle \frac{\ln <!--\; \; -->(1+e^{\mathrm{\alpha <!--\; \alpha \; -->}x})}{\ln <!--\; \; -->(1+e^{\mathrm{\beta <!--\; \beta \; -->}x})}}$?
Where $\alpha >0$ and $\beta >0$

Calculus - limit of a function with logarithms
Compute
$\underset{x\to <!--\; \to \; -->0_{+}}{lim}\ln <!--\; \; -->(x\ln <!--\; \; -->a)\ln <!--\; \; -->\left(\frac{\ln <!--\; \; -->(ax)}{\ln <!--\; \; -->(\frac{x}{a})}\right)$
where $a>1$
I am trying to get to a result withou using any advanced methods or even things such as l'Hospital's rule etc..
I got to a phase where I took the limit of the first logarithm which we can see tends to 0 from rewriting it as $\ln <!--\; \; -->a^x$. Then I wanted to make some adjustments to the second part of the expression and I got to the stage where I have the limit of
$(\ln <!--\; \; -->a^2)\left(\frac{x-<!--\; -\; -->a}{a}\right)$
That wont give me exact result but I should be able to justify that it the expression is defined and by multiplying it with the first limit which is $0$, the result should also be $0$
Can somebody please tell me how correct or wrong I am? Thanks.

solve logarithmic equation without numerical methods
Is there algebraic method to solve following equation for x:
$ax+b\ln <!--\; \; -->x+c=0$
with $a,b,c$ constants without using numerical methods and ln means natural logarithm.

Lambert- W -Function calculation?
I have an equation of the form:
$n\log <!--\; \; -->n=x$
Upon searching I came across the term "Lambert- W -Function" but couldn't find a proper method for evaluation, and neither any computer code for it's evaluation.
Any ideas as to how I can evaluate?

When can I use the natural log to help solve an integral?
Why is it okay to do this: $\int <!--\; \int \; -->\frac{1}{x-<!--\; -\; -->2}dx=\ln <!--\; \; -->(x-<!--\; -\; -->2)$
but not this: $\int <!--\; \int \; -->\frac{1}{1-<!--\; -\; -->x^2}dx=\ln <!--\; \; -->(1-<!--\; -\; -->x^2)$

What does ${\log }_a<!--\; \; -->(b)$ equal to?
Does
${\log }_a<!--\; \; -->(b)=\frac{{\log }_c<!--\; \; -->(b)}{{\log }_c<!--\; \; -->(a)}$
or
${\log }_a<!--\; \; -->(b)=\frac{\ln <!--\; \; -->(b)}{\ln <!--\; \; -->(a)}$
??
Is there any difference between the two?

Properties of Logarithms
How do you simplify the following expression?
$\log <!--\; \; -->\left(3^{(5^7)}\right)$
I know that logarithms are like the inverse of exponents, but are there any tricks to simplify powers inside logarithms?
Edit: There are five answer choices:
(A) $7\log <!--\; \; -->(3^5)$
(B) $35\log <!--\; \; -->(3)$
(C) $7(\log <!--\; \; -->(3))(\log <!--\; \; -->(5))$
(D) $7(\log <!--\; \; -->(3)+\log <!--\; \; -->(5))$
(E) $5^7\log <!--\; \; -->(3)$

Basic Logarithm Equation
${\log }_2<!--\; \; -->(x)={\log }_x<!--\; \; -->(2)$
Using the change of base theorem: ${\displaystyle \frac{\log <!--\; \; -->(x)}{\log <!--\; \; -->(2)}}={\displaystyle \frac{\log <!--\; \; -->(2)}{\log <!--\; \; -->(x)}}$
Multiplied the denominators on both sides: $\log <!--\; \; -->(x)\log <!--\; \; -->(x)=\log <!--\; \; -->(2)\log <!--\; \; -->(2)$
I kind of get stuck here. I know that you can't take the square root of both sides of the equation, but still, $x=2$ seems to be an obvious solution to the equation. I've missed $2^{-<!--\; -\; -->1}$ or $\frac{1}{2}$ as another answer to the equation, which I am struggling to get to.
Any help will be greatly appreciated, thanks in advance.

Homework solving exponential equation, logarithmic equation and exponential equation.
I need help with three homework questions.¨
First one:
$\sqrt{3x^2-<!--\; -\; -->2x-<!--\; -\; -->15}=x+1$
I don't know how to get the right answer. The answer is supposed to be 4. I get:
$3x^2-<!--\; -\; -->2x-<!--\; -\; -->15=(x+1{)}^2$
$3x^2-<!--\; -\; -->2x-<!--\; -\; -->15=x^2+2x+1$
$2x^2-<!--\; -\; -->4x-<!--\; -\; -->14=0$
$x^2-<!--\; -\; -->2x-<!--\; -\; -->7=0$
$x_1=1+\sqrt{8}$
$x_2=1-<!--\; -\; -->\sqrt{8}$
Second question is:
$ln(x-<!--\; -\; -->7)-<!--\; -\; -->2ln(x-<!--\; -\; -->1)=-<!--\; -\; -->2ln5$
where i come to this stage but don't know how to solve it
$\frac{x-<!--\; -\; -->7}{(x-<!--\; -\; -->1{)}^2}=\frac{1}{25}$
the answer is supposed to be $x=11$ or $x=16$
Third question is
$9^x-<!--\; -\; -->4\ast <!--\; \ast \; -->3^x+3=0$
I have come this far
$3^{2x}-<!--\; -\; -->4\ast <!--\; \ast \; -->3^x+3=0$
answer is supposed to be $x=1$ or $x=0$

Can $\log <!--\; \; -->(x)\log <!--\; \; -->(y)$ be reduced?
I'm currently taking Pre-Calc and am learning about logs. I know that $\log <!--\; \; -->(xy)=\log <!--\; \; -->(x)+\log <!--\; \; -->(y)$, but can $\log <!--\; \; -->(x)\log <!--\; \; -->(y)$ be reduced further?

Find number of digits of a number in another base
How can I solve this question:
Given that $\log <!--\; \; -->3$ is about $0.48$, approximately how many digits are in the number ${10}^{150}$ if it were written in base $3$
Thanks!

How to find the limit $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\ln <!--\; \; -->(x^2+x+5)}{\ln <!--\; \; -->(x^8-<!--\; -\; -->x+3)}$?
I tried reducing it to the form $lim(1+x_n{)}^{1/x_n}=1$, but that didn't work.

Simplifying trivial expression
How to simplify an expression: $e^{\frac{-<!--\; -\; -->{\ln }^2<!--\; \; -->y}{2}}$? I mean because as if $e^{\ln <!--\; \; -->y}=y$ then my guess is that this could be simplified in a similar manner thus how to do that?

${\ln }^2<!--\; \; -->(x)\stackrel{?}{=}2\ln <!--\; \; -->(x)$
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?