Learn How to Graph Logarithmic Equations

How do I solve this logarithmic equation? solving for x
How do I solve this logarithmic equation?
${\log }_4<!--\; \; -->x=({\log }_4<!--\; \; -->x{)}^{1/3}$

How to prove if log is rational/irrational
I'm an English major, now doubling in computer science. The first course I'm taking is Discrete Mathematics for Computer Science, using the MIT 6.042 textbook.
Within the first chapter of the book's practice problems, they ask us multiple times to prove that some log function is either rational or irrational.
Specific cases make more sense than others, but I would really appreciate any advice on how to approach these problems. Not how to carry them out algebraically, but what thought constructs are necessary to consider a log being (ir)rational.
For example, in the case of ${\sqrt{2}}^{2{\log }_2<!--\; \; -->3}$, proving that $2{\log }_2<!--\; \; -->3$ is irrational (and therefore $a^b$, when $a=\sqrt{2}$ and $b=2{\log }_2<!--\; \; -->3$ is rational) is not an easily solvable problem. I understand the methods of proofs, but the rules of logs are not intuitive to me.
A section from my TF's solution is not something I would know myself to construct:
Since $2<3$ we know that ${\log }_2<!--\; \; -->3$ is positive (specifically it is greater than 1), and hence so is $2{\log }_2<!--\; \; -->3$. Therefore, we can assume that a and b are two positive integers. Now $2{\log }_2<!--\; \; -->3=a/b$ implies $2^{2{\log }_2<!--\; \; -->3}=2^{a/b}$. Thus
$2^{a/b}=2^{2{\log }_2<!--\; \; -->3}=2^{{\log }_2<!--\; \; -->3^2}=3^2=9\text{,}$
and hence $2^a=9^b$
Any advice on approaching thought construct to logs would be greatly appreciated!

Integral ${\int <!--\; \int \; -->}_0^1\frac{\ln <!--\; \; -->x}{x^2+1}\cdot <!--\; \cdot \; -->\ln <!--\; \; -->\left(\frac{3x^2+1}{x^2+3}\right)dx$
I need to evaluate the following integral:
${\int <!--\; \int \; -->}_0^1\frac{\ln <!--\; \; -->x}{x^2+1}\cdot <!--\; \cdot \; -->\ln <!--\; \; -->\left(\frac{3x^2+1}{x^2+3}\right)dx.$
Could you suggest how to find a closed form for it? I am not sure if there is one, but the integrand seems simple enough, so I hope it might exist.

logarithms equations,
Hello I have following problem: solve equation $\log <!--\; \; -->(x-<!--\; -\; -->5{)}^2+\log <!--\; \; -->(x+6{)}^2=2$and I rewrited this equation as
$2\log <!--\; \; -->(x-<!--\; -\; -->5)+2\log <!--\; \; -->(x+6)=\log <!--\; \; -->100⟹<!--\; ⟹\; -->2(\log <!--\; \; -->(x-<!--\; -\; -->5)(x+6))=\log <!--\; \; -->100⟹<!--\; ⟹\; -->\log <!--\; \; -->x^2+x-<!--\; -\; -->30=\log <!--\; \; -->10⟹<!--\; ⟹\; -->x^2+x-<!--\; -\; -->40=0$
and I solved this equation, but I obtained only two solutions and there should be four, so I wonder if it is necessary to create $\log <!--\; \; -->(x-<!--\; -\; -->5{)}^2(x+6{)}^2=\log <!--\; \; -->100$ or is a simplier way.

Which is bigger $x^{log(x)}$ or $(log(x){)}^x$
I'm trying to find out which is bigger $x^{log(x)}$ or $(log(x){)}^x$ as $x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$
I tried to take the log of both but I didnt't reach any where.

Finding the inverse of a natural log
How would I find the inverse of
$\ln <!--\; \; -->(8x-<!--\; -\; -->64)?$
I've tried put $8x-<!--\; -\; -->64$ as the power to the base of $e$, I don't know what to do from there on, thanks in advance

Deriviative of natural log help finding
$y=7\ln <!--\; \; -->\frac{11}{x}$
I need to use the product rule please
$\frac{d}{dx}(7)\cdot <!--\; \cdot \; -->(\ln <!--\; \; -->(11/x)+\frac{d}{dx}(\ln <!--\; \; -->\frac{11}{x})\cdot <!--\; \cdot \; -->7$
then
$0+\frac{d}{dx}\frac{11}{x}\cdot <!--\; \cdot \; -->7$
what do i do

Why does the value $-<!--\; -\; -->\ln <!--\; \; -->(|\cos <!--\; \; -->(x)|)$ become $\ln <!--\; \; -->(|\sec <!--\; \; -->(x)|)?$
I was doing an integral and I got my final answer as that, but I don't understand how you can just send the negative sign inside and make it $\sec <!--\; \; -->(x).$

$100(1.2{)}^t=\mathrm{\alpha <!--\; \alpha \; -->}t$. How do I solve for t?

Logarithm rules for complex numbers
Are the logarithm rules true for complex numbers?
We know that for positive real numbers $a$, $b$, $c$ and real number $d$ that:
${\log }_b<!--\; \; -->\left(a^d\right)=d{\log }_b<!--\; \; -->(a)$
${\log }_b<!--\; \; -->(a)=\frac{{\log }_c<!--\; \; -->(a)}{{\log }_c<!--\; \; -->(b)}$
${\log }_b<!--\; \; -->(xy)={\log }_b<!--\; \; -->(x)+{\log }_b<!--\; \; -->(y)$
${\log }_b<!--\; \; -->\left(\frac{x}{y}\right)={\log }_b<!--\; \; -->(x)-<!--\; -\; -->{\log }_b<!--\; \; -->(y)$
We also know that
${\log }_b<!--\; \; -->\left(b^d\right)=d$
Does this extend to complex numbers as $a$, $b$, $c$ or $d$?
My instinct is that
${\log }_b<!--\; \; -->\left(b^{s+t\mathrm{i}}\right)=s+t\mathrm{i}$
In other words, I'm pretty confident that the last formula works when $d$ is a complex number

Solve $\frac{\log <!--\; \; -->(2x+1)-<!--\; -\; -->\log <!--\; \; -->4}{1-<!--\; -\; -->\log <!--\; \; -->(3x+2)}=1$
My attempt:
$\frac{\log <!--\; \; -->(2x+1)-<!--\; -\; -->\log <!--\; \; -->4}{1-<!--\; -\; -->\log <!--\; \; -->(3x+2)}=1$
$\frac{\log <!--\; \; -->(2x+1)-<!--\; -\; -->\log <!--\; \; -->4}{\log <!--\; \; -->10-<!--\; -\; -->\log <!--\; \; -->(3x+2)}=\log <!--\; \; -->10$
$\frac{\log <!--\; \; -->\frac{(2x+1)}{4}}{\log <!--\; \; -->\frac{10}{(3x+2)}}=\log <!--\; \; -->10$
$\frac{\frac{(2x+1)}{4}}{\frac{10}{(3x+2)}}=10$
$\frac{(2x+1)}{4}=10\cdot <!--\; \cdot \; -->\frac{10}{(3x+2)}$
(is this even correct ??)
$\frac{(2x+1)}{4}=\frac{100}{(3x+2)}$
$(2x+1)100=4(3x+2)$
$200x+100=12x+8$
$188x=-<!--\; -\; -->92$
$x=-<!--\; -\; -->23/47$
The solution is 2.

Value of $2^{-<!--\; -\; -->1-<!--\; -\; -->i}$ in the complex
I am trying to find the value of $2^{-<!--\; -\; -->1-<!--\; -\; -->i}$
I rewrite it like this, $2^{-<!--\; -\; -->1-<!--\; -\; -->i}=e^{\ln <!--\; \; -->(2)(-<!--\; -\; -->1-<!--\; -\; -->i)}=\frac{1}{e^{\ln <!--\; \; -->2}{e}^{i\ln <!--\; \; -->2}}=1/2$
Since $e^{i\ln <!--\; \; -->2}=e^{Re(i\ln <!--\; \; -->2)}=e^0=1$
This looks way nicer than it should be, I think, can anyone tell me where I go wrong, and maybe a good way to do this problem?
Thank you for any input.

Find $\underset{x\to <!--\; \to \; -->0}{lim}\left(\frac{\ln <!--\; \; -->(\cos <!--\; \; -->x)}{x\sqrt{1+x}-<!--\; -\; -->x}\right)$ efficiently
Now, it looked to me like a classic L'Hôpital's rule case. Indeed, I used it (twice), but then things became messy and complicated.
Am I missing the point of this exercise? I mean, there must be a "nicer" way. Or should I stick with this road?
EDIT:
Regarding Yiorgos's answer: Why is the following true?
$\ln <!--\; \; -->(1-<!--\; -\; -->\frac{x^2}{2})\approx <!--\; \approx \; -->-<!--\; -\; -->\frac{x^2}{2}$

Prove that for all $x>0$, $1+2\ln <!--\; \; -->x\le <!--\; \le \; -->x^2$
Prove that for all $x>0$
$1+2\ln <!--\; \; -->x\le <!--\; \le \; -->x^2$
How can one prove that?

logarithmic function between two points
I need to find the logarithmic curve between two points
$A(0,5),B(180,9)$
We know that the formula for logarithmic function is: $f(x)=\log <!--\; \; -->(x)$ so
$5=\log <!--\; \; -->(0),9=\log <!--\; \; -->(180)$
But that's impossible because $\log <!--\; \; -->(0)$ is undefined. What Did I do wrong?
Following the below advice I'm still stuck
$a^5=0-<!--\; -\; -->b$ and $a^9=180-<!--\; -\; -->b$
then
$a^9=180+a^5$
$a^4=180$
$a=3.66$
Now let's plug a in our original formula
${3.66}^5=-<!--\; -\; -->b$
$b=-<!--\; -\; -->656.7$
$f(x)=\log <!--\; \; -->3.66(x+656.7)$
I did a little bit of fiddling with a graph and at the end of the day what I was looking for was
$f(x)=\log <!--\; \; -->1.77(x-<!--\; -\; -->5)$
I would be awesome to understand how to achieve this result without playing randomly with excel.

How to prove the following inequality of logarithm?
Let $x,y,z\in <!--\; \in \; -->\mathbb{C}.$ Suppose
$z=\frac{1}{2}(xy\pm <!--\; \pm \; -->\sqrt{x^2{y}^2-<!--\; -\; -->4(x^2+y^2)}).$
Show that
$lo{g}^{+}|z|\le <!--\; \le \; -->lo{g}^{+}|x|+lo{g}^{+}|y|+log2.$
Where $lo{g}^{+}\mathrm{\varphi <!--\; \varphi \; -->}=max\{0,log\mathrm{\varphi <!--\; \varphi \; -->}\}.$
Here we are also considering the complex root function with respect to the principle branch of logarithm.

Why is it trivial that ${(1+\frac{2\ln <!--\; \; -->3}{3})}^{-<!--\; -\; -->3/2}\le <!--\; \le \; -->\frac{2}{3}$
Can someone tell me why
${(1+{\displaystyle \frac{2\ln <!--\; \; -->3}{3}})}^{-<!--\; -\; -->3/2}\le <!--\; \le \; -->{\displaystyle \frac{2}{3}}$
is trivial because for me its not and I will need to do the calculation to see it.

absolute value of logarithm
I have a problem with understand how function $2^{|{\log }_{1/2}<!--\; \; -->x|}$ obtains values for the negative x ? I thought that there is the assumption that x>0 but wolframalpha shows chart that for negative x also obtains values.
I tried to do it in this way: $2^{|{\log }_{1/2}<!--\; \; -->x|}$ for $x\in <!--\; \in \; -->(0,1)$ have formula $y=\frac{1}{x}$ and for $x\in <!--\; \in \; -->[1,+\mathrm{\infty <!--\; \infty \; -->})$ equals $y=x$
But how it looks for the negative values?

From $\underset{\lfloor <!--\; \lfloor \; -->\log <!--\; \; -->n\rfloor <!--\; \rfloor \; -->+1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}n/2^r$ to $\underset{r=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}1/2^r$
$E[h]=E[\underset{r=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{I}_r]=\underset{r=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}E[I_r]$
$=\underset{r=1}{\overset{\lfloor <!--\; \lfloor \; -->\log <!--\; \; -->n\rfloor <!--\; \rfloor \; -->}{\sum <!--\; \sum \; -->}}E[I_r]+\underset{\lfloor <!--\; \lfloor \; -->\log <!--\; \; -->n\rfloor <!--\; \rfloor \; -->+1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}E[I_r]$
$\le <!--\; \le \; -->\underset{r=1}{\overset{\lfloor <!--\; \lfloor \; -->\log <!--\; \; -->n\rfloor <!--\; \rfloor \; -->}{\sum <!--\; \sum \; -->}}1+\underset{\lfloor <!--\; \lfloor \; -->\log <!--\; \; -->n\rfloor <!--\; \rfloor \; -->+1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}n/2^r$
$\le <!--\; \le \; -->\log <!--\; \; -->n+\underset{r=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}1/2^r$
$=\log <!--\; \; -->n+2$
I'm trying to understand how you go from $\underset{\lfloor <!--\; \lfloor \; -->\log <!--\; \; -->n\rfloor <!--\; \rfloor \; -->+1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}n/2^r$ to $\underset{r=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}1/2^r$

logarithm of a complex number?
I have a task to study a function like this one:
$F(z)=\frac{\ln <!--\; \; -->(e^{i{z}^4})}{z^3}$
I'm trying to simplify this: since the exponential is the inverse function of $\ln <!--\; \; -->()$ can we simplify it to be:
$iz$
$\log <!--\; \; -->(z)=ln(r)+i\mathrm{\theta <!--\; \theta \; -->}$
But my problem is with :
$\ln <!--\; \; -->(e^{iz})=\ln <!--\; \; -->(e^{ix-<!--\; -\; -->y})$ can we simplify it to: $ix-<!--\; -\; -->y$?