Learn How to Graph Logarithmic Equations

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?

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?

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?

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.

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.

$\frac{\ln <!--\; \; -->(x^2)}{\ln <!--\; \; -->(x)}=2$? Why?
Upon trying to evaluate $\frac{\ln <!--\; \; -->(x^2)}{\ln <!--\; \; -->(x)}$, i've found that google plots it as always equal to 2, other than 0 where it is undefined. Why is this the case?

If $x^y=y^x$ $(x,y\in <!--\; \in \; -->R,x,y>0,x\ne <!--\; \ne \; -->0)$ and $x^p=y^q$ $(p,q\in <!--\; \in \; -->R/\{0\},p\ne <!--\; \ne \; -->q)$, then product xy is equal to?
Solution for this one is $(\frac{p}{q}{)}^{\frac{p+q}{p-<!--\; -\; -->q}}$ , but I do not understand how I am supposed to get here, I guess something with logarithms but not sure what?

How to differentiate $(\ln <!--\; \; -->n{)}^{\ln <!--\; \; -->n}$?
$f(n)=(\ln <!--\; \; -->n{)}^{\ln <!--\; \; -->n}$
Can someone explain me how to differentiate the above function?
I am trying the following solution
$f^{\prime }(n)=\frac{\ln <!--\; \; -->n((\ln <!--\; \; -->n{)}^{\ln <!--\; \; -->n}-<!--\; -\; -->1)}{(\frac{1}{n}\ln <!--\; \; -->2)}$

Find x, if ${\log }_{15}<!--\; \; -->{\left(\frac{2}{9}\right)}^{}={\log }_3<!--\; \; -->\left(x\right)={\log }_5<!--\; \; -->(1-<!--\; -\; -->x)$
I tried switching everything to base 15, but that didn't work out well.

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.

Initial value problem through origin
$\frac{dz}{dt}=8t\ast <!--\; \ast \; -->e^z$, Through the origin
I have never done an initial value problem before, but I took it to mean that it gave me the initial value of the differential equation (0, 0) and that I should find the general value of the differential equation from it. Please correct me if I am wrong about this.
I started off by separating the variables:
$\frac{dz}{e^z}=8t\ast <!--\; \ast \; -->dt$
I then integrated:
$-<!--\; -\; -->e^{-<!--\; -\; -->z}=4t^2+c$
After dividing by -1 and taking the natural log I have:
$z=-<!--\; -\; -->ln(-<!--\; -\; -->4t^2+c)$
Here is where my problem is, no matter what I plug in for c I cannot get the function to equal 0 when t = 0. Even for another value of t it does not seem to possible to get 0 with this differential equation.

Is this logarithmic inequality true?
Assume we have two complex variables $h_i$ and $h_d$ which satisfy the following relationship
$2|h_i{|}^2\le <!--\; \le \; -->|h_d{|}^2$
can we say that
$\log <!--\; \; -->(1+\frac{||h_d|-<!--\; -\; -->h_i{|}^2}{2})\le <!--\; \le \; -->\log <!--\; \; -->(1+(1+\frac{1}{\sqrt{2}}{)}^2\frac{|h_d{|}^2}{2})?$
Many thanks

Limit of the sine and logarithm function
How do I calculate the following?
$\underset{x\to <!--\; \to \; -->0^{+}}{lim}(-<!--\; -\; -->\ln <!--\; \; -->(x)\sin <!--\; \; -->(x))$

How to solve $\ln <!--\; \; -->(x)=2x$
I know this question might be an easy one. but it has been so long since I solved such questions and I didn't find a an explanation on the internet. I'd like if someone can remind me.
I reached that $e^{2x}=x$, but didn't know how to continue from here. I remember something that has to do with bases and equalizing parameters, but how do I do that in this case?

Logarithmic inequality: ${\log }_{1/3}^2<!--\; \; -->(x^2-<!--\; -\; -->3x+2)-<!--\; -\; -->{\log }_{1/3}<!--\; \; -->(x-<!--\; -\; -->1)>{\log }_{1/3}<!--\; \; -->(x-<!--\; -\; -->2)+6$

Express this logarithm in terms of a and b
How do I express ${\log }_5<!--\; \; -->2$ in terms of $a$ and $b$ if:
${\log }_6<!--\; \; -->2=a$ and ${\log }_5<!--\; \; -->3=b$
I've tried:
Converting the a and b equations to fractions, and substituting $\log <!--\; \; -->2$ and $\log <!--\; \; -->5$ with $a\log <!--\; \; -->6$ and $(\log <!--\; \; -->3)/b$ respectively, but I ended up with the same equation after simplifying things out.
I've also tried to convert the $2$ in ${\log }_5<!--\; \; -->2$ into fractions and go from there, but I went up going in circles and never got anywhere.
How would I solve this question?

How to derive the value of $\log <!--\; \; -->(-<!--\; -\; -->1)$?
My book gives the following relation which I cannot derive myself. How to approach it?
$\log <!--\; \; -->(-<!--\; -\; -->1)=(2n+1)\mathrm{\pi <!--\; \pi \; -->}i,n=0,\pm <!--\; \pm \; -->1,...$
The definition of logarithm I am using is, $\log <!--\; \; -->z$ is any complex number $w$ such that $e^w=z$

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.