Learn How to Graph Logarithmic Equations

Calculate $I={\int <!--\; \int \; -->}_1^e\frac{(1+\ln <!--\; \; -->x)x}{(1+x\ln <!--\; \; -->x{)}^2}dx$
Please help me solve this: (level = high school)
Thanks:)

What is the difference between log scale. and plotting logarithms?
I find plots in scientific literature beyond confusing. I understand quite clearly the difference between a linear and a logarithmic scale, and when each is desirable. Suppose we are plotting values for the equation
$$y=f(x)$$
If the points in x vary over several orders of magnitude, a log scale is useful to capture all the points. However, quite often a log scale will be used, and the x axis will be labeled as $\log <!--\; \; -->(x)$, particularly in engineering literature.
Is this even correct, or just sloppy notation to indicate that a $\log$ is being used? For instance, if I read a value on the scale itself, is this not representing just x, instead of $\log <!--\; \; -->(x)$?
Finally, how does plotting $\log <!--\; \; -->(x)$, and keeping the scale linear, differ from plotting the raw value of x on a logarithmic scale? In the first case, to understand the data visually, I have to somehow be able to exponentiate each x value in my mind, while in the second case there is no work to be done.
Finally, a common phrase when using a $\log$ scale is "the paper is taking the log". What does this even mean?

Solving logarithmic equation $2\log <!--\; \; -->(x)+1=\log <!--\; \; -->(19x+2)$
I know the solution has to be $x=2$
However I can't find the manual steps (Wolfram doesn't know the manual steps either).
This is all I got
$$\log <!--\; \; -->(x^2)+1=\log <!--\; \; -->(19x+2)$$
$$\log <!--\; \; -->(x^2)-<!--\; -\; -->\log <!--\; \; -->(19x+2)=-<!--\; -\; -->1$$
$$\log <!--\; \; -->\left(\frac{x^2}{19x+2}\right)=-<!--\; -\; -->1$$

Write the following as a sum of logarithms:
$$\log <!--\; \; -->(\sqrt[5]{\frac{x^6{y}^{10}}{z^7}})$$

How to solve this Diophantine equation (involving natural logarithms)?
The equation is $r=\ln <!--\; \; -->a+b\ln <!--\; \; -->c$ where $r\in <!--\; \in \; -->\mathbb{R}$ is fixed and $a,b,c\in <!--\; \in \; -->\mathbb{N}$. In other words, for arbitrary real r, how can one say whether a solution (in form above) exists or not.

Indefinite integrals with natural logs
I know the integral of $\frac{1}{x}$ is $\log <!--\; \; -->(x)$ but I'm not sure how to solve this problem, any help would be appreciated:
$${\int <!--\; \int \; -->}_2^3\frac{1}{x\ln <!--\; \; -->x}$$
I think I need to substitute $x\ln <!--\; \; -->x$ but from there I get stuck.

Given that ${\log }_9<!--\; \; -->p={\log }_{12}<!--\; \; -->q={\log }_{16}<!--\; \; -->(p+q)$ find the value of $q/p$
This is not homework, it's just a brain teaser which I can't solve, just some hints should be sufficient, I know that from this we get:
$$(1/4){\log }_2<!--\; \; -->(p+q)=(1/2){\log }_3<!--\; \; -->p=\frac{{\log }_3<!--\; \; -->q}{1+2{\log }_3<!--\; \; -->2}$$
now I'd like to combine these quantities in some way so I can see the value of $q/p$ from this, but I can't seem to be able to figure it out.

Different results when integrating $1/(x\ln <!--\; \; -->(x))$ partially/by substitution.
By substitution I get $ln(ln(x))$. Partially something completely different:
$$\int <!--\; \int \; -->\frac{1}{x\ln <!--\; \; -->(x)}=\int <!--\; \int \; -->\frac{1}{x}\frac{1}{\ln <!--\; \; -->(x)}dx=\frac{\ln <!--\; \; -->(x)}{\ln <!--\; \; -->(x)}-<!--\; -\; -->\int <!--\; \int \; -->-<!--\; -\; -->\frac{1}{x\ln <!--\; \; -->(x{)}^2}dx$$
$$\int <!--\; \int \; -->\frac{1}{x\ln <!--\; \; -->(x)}dx=1+C+\int <!--\; \int \; -->\frac{1}{x\ln <!--\; \; -->(x)}dx$$
Provided that this is correct, it doesn't really solve the integral - why is that?
Side info: this is not homework.

Prove that $-<!--\; -\; -->(p_1+p_2)\log <!--\; \; -->p_1+p_2\le <!--\; \le \; -->-<!--\; -\; -->p_1\log <!--\; \; -->p_1-<!--\; -\; -->p_2\log <!--\; \; -->p_2$ provided that $p_1,p_2>0$
WTS:
$$-<!--\; -\; -->(p_1+p_2)\log <!--\; \; -->(p_1+p_2)\le <!--\; \le \; -->-<!--\; -\; -->p_1\log <!--\; \; -->p_1-<!--\; -\; -->p_2\log <!--\; \; -->p_2\mathrm{\forall <!--\; \forall \; -->}{p}_1,p_2>0$$
Any hints on this? I've tried to set it up as a proof by contradiction, and jiggled around with the terms both as logarithms and after exponentiating everything, but no luck.

Prove $x^n<n^n{2}^x$
Given that
$$x<2^x$$
is always true, use it to prove that
$$x^n<n^n{2}^x$$
Here are the steps that I've taken so far:
Reduce
$$x<2^x$$
to
$$\log <!--\; \; -->(x)<x$$
Then
$$x^n<n^n{2}^x$$
$$n\log <!--\; \; -->(x)<n\log <!--\; \; -->(n)+x$$
$$\log <!--\; \; -->(x)<\log <!--\; \; -->(n)+\frac{x}{n}$$
And that's where I got stuck.
If it were
$$\log <!--\; \; -->(x)<\log <!--\; \; -->(n)+x$$
the proof would be self evident, because since
$$\log <!--\; \; -->(x)<x$$
it would be obvious that $(x+\text{anything else})$ would be more than $\log <!--\; \; -->(x)$, but in this case it isn't.
Is there something else that I'm missing? By direct calculation my last step would be true, but how do I relate it to the first statement?
Any help would be greatly appreciated, guys. Thanks a lot!

How to get the graph for $y={\log }_{1/a}<!--\; \; -->(x)$ from $y={\log }_a<!--\; \; -->(x)$?
I know that it is symmetric the Ox axle, but I can't prove it.

$c=(a^x-<!--\; -\; -->b^x)$ where $a$,$b$ and $c$ are known real constants. Solve for $x$
I tried taking $\log$ on both side but i ended with $\log <!--\; \; -->(a^x-<!--\; -\; -->b^x)$ which is difficult to solve. Does anybody has idea how to solve the above equation for $x$

How to show that $n^{\ln <!--\; \; -->(\ln <!--\; \; -->(n))}=\ln <!--\; \; -->(n{)}^{\ln <!--\; \; -->(n)}$
I have verified that $n^{\ln <!--\; \; -->(\ln <!--\; \; -->(n))}=\ln <!--\; \; -->(n{)}^{\ln <!--\; \; -->(n)}$ by plugging in values for $n$, but do not understand why it is true. I am not aware of any $\log$ rules that can be used to simplify $n^{\ln <!--\; \; -->(\ln <!--\; \; -->(n))}$

How many digits will $5^{4^{3^2}}$ produce?
According to Wolfram Alpha, $5^{4^{3^2}}$ evaluates to an integer with $183231$ digits. How does one find out how many digits such a calculation will produce?

Is $\ln <!--\; \; -->(x^{p(x)})=p(x)\ln <!--\; \; -->(x)$?
I am trying to prove that:
$x^{\frac{\ln <!--\; \; -->(\ln <!--\; \; -->(x))}{\ln <!--\; \; -->(x)}}=\ln <!--\; \; -->(x)$
My "solution":
$e^{\ln <!--\; \; -->\left(x^{\frac{\ln <!--\; \; -->(\ln <!--\; \; -->(x))}{\ln <!--\; \; -->(x)}}\right)}=e^{\frac{\ln <!--\; \; -->(\ln <!--\; \; -->(x))}{\ln <!--\; \; -->(x)}\ln <!--\; \; -->(x)}=e^{\ln <!--\; \; -->(\ln <!--\; \; -->(x))}=\ln <!--\; \; -->(x)$
Is the first step valid, i.e is $\ln <!--\; \; -->(x^{p(x)})=p(x)\ln <!--\; \; -->(x)$
How can I find out for myself?

How to prove that $f(x)-<!--\; -\; -->f(x-<!--\; -\; -->1)$ approaches $\frac{{\log }_{10}<!--\; \; -->(10)}{{\log }_{10}<!--\; \; -->(e)}$?
Let
$$f(x)=\underset{n=1}{\overset{{10}^x}{\sum <!--\; \sum \; -->}}\frac{1}{n}$$
I noticed that as x approaches $\mathrm{\infty <!--\; \infty \; -->}$,
$f(x)-<!--\; -\; -->f(x-<!--\; -\; -->1)\approx <!--\; \approx \; -->2.3025$. After a bit of experimenting, I found that $\mathrm{2.3025...}=\frac{{\log }_{10}<!--\; \; -->(10)}{{\log }_{10}<!--\; \; -->(e)}$
How can I prove that as x approaches $\mathrm{\infty <!--\; \infty \; -->}$, $f(x)-<!--\; -\; -->f(x-<!--\; -\; -->1)$ approaches $\frac{{\log }_{10}<!--\; \; -->(10)}{{\log }_{10}<!--\; \; -->(e)}$?

Simplifying the exponential expression $e^{-<!--\; -\; -->4\ln <!--\; \; -->x+8\ln <!--\; \; -->y+2}$

An efficient technique to test if an exponential of logs gives an integer
Is there an efficient way of testing if the resulting value of an exponential gives an integer without actually expanding the equation.
For example:
$\log <!--\; \; -->(12)-<!--\; -\; -->\log <!--\; \; -->(4)=1.09861\dots <!--\; \dots \; -->$
and is a transcendental number.
But if we take the exponential, we get an integer:
$e^{\log <!--\; \; -->(12)-<!--\; -\; -->\log <!--\; \; -->(4)}=e^{1.09861\dots <!--\; \dots \; -->}=3$
This can be easily shown by proving that the equation is equivalent to $12/3$
However is there a way to prove this while staying in the logarithm nation... We are assuming that the two numbers $12$ and $4$ are very large...
Can I find say numerical stability bounds to this simple equation:
${2.71828}^{2.48491-<!--\; -\; -->1.38629}=3\dots <!--\; \dots \; -->$
such that if I use $n$ digits, I will get an integer close the the $n$-th digit.
This may look trivial, but let's assume we do not have say $\log <!--\; \; -->(12)$, but only $2.48491\dots <!--\; \dots \; -->$ up to some precision of k-digits. Same thing for $\log <!--\; \; -->(4)$, etc.

Logs - Simplifying with arbitrary constant
I've tried simplifying my answer, which has a constant in it. I would like to know if I am on the right track:
$$\ln <!--\; \; -->(y)=-<!--\; -\; -->\frac{x^2}{2y^2}+C$$
C can be considered as an Arbitrary Constant. From this we can say that:
$$\ln <!--\; \; -->(k)=C$$
We can further simplify the above to the following:
$$y=k.e^{\frac{-<!--\; -\; -->x^2}{2y^2}}$$
Would this be correct?

Finding Stationary Points of Natural Log Function
$$f(x)=x-<!--\; -\; -->2\ln <!--\; \; -->(x^2+3)$$
I started by using the chain rule on $x^2+3$ which gives me $\frac{2x}{x^2}+3$
At this point I tried to multiple $\frac{2x}{x^2}+3$ by $x-<!--\; -\; -->2$ - is this correct? How can I then find the stationary point?
I understand that you may not wish to give me the answer on a silver platter. Perhaps somebody could kindly provide me with an example in a similar format and the steps involved in solving it.