Learn How to Graph Logarithmic Equations

How do you solve logarithmic equations like this one?
How do you solve
$3\log <!--\; \; -->(x-<!--\; -\; -->15)={\left(\frac{1}{4}\right)}^x?$
The solution is approximately 16. How would you solve a logarithmic equation with an solution approximately equal to a number without using a graphing calculator?

Solving $7^{2x}\cdot <!--\; \cdot \; -->4^{x-<!--\; -\; -->2}={11}^x$
I am trying to solve this equation but I am stuck little bit right now. This is how I did it:
$$\begin{gathered} 7^{2x}\cdot <!--\; \cdot \; -->4^{x-<!--\; -\; -->2}={11}^x \\ \text{log both sides} \\ \log <!--\; \; -->(7^{2x}\cdot <!--\; \cdot \; -->4^{x-<!--\; -\; -->2})=\log <!--\; \; -->({11}^x) \\ \log <!--\; \; -->(7^{2x})+\log <!--\; \; -->(4^{x-<!--\; -\; -->2})=\log <!--\; \; -->({11}^x) \\ 2x\log <!--\; \; -->(7)+(x-<!--\; -\; -->2)\log <!--\; \; -->(4)=x\log <!--\; \; -->(11) \\ \text{I got this far} \end{gathered}$$
Can somebody give me idea, am I on right track, and if I am, how can I solve it further. Thanks.

How would you fin the integer closest to
log(log(1234567891011121314...2013))
where the number is the concatenation of numbers 1 through 2013 inclusive. log() in this case is log base 10.
Also, how would you find the remainder when it is divided by 75?

Solving equations of type $x^{1/n}={\log }_n<!--\; \; -->x$
First, I'm a new person on this site, so please correct me if I'm asking the question in a wrong way.
I thought I'm not a big fan of maths, but recently I've stumbled upon one interesting fact, which I'm trying to find an explanation for. I've noticed that graphs of functions $y=x^{1/n}$ and $y={\log }_n<!--\; \; -->x$ , where $n$ is given and equal for both functions, always have $2$ intersection points. This means, equation $x^{1/n}={\log }_n<!--\; \; -->x$ must have $2$ solutions, at least it's what I see from the graphs.
I've tried to solve this equation analytically for some given $n$, like $4$, but my skills are very rusty, and I cannot come up with anything. So I'm here for help, and my question(-s) are:
are these $2$ functions always have $2$ intersection points?
if yes, why, if not, when not?
how to solve equations like $x^{1/n}={\log }_n<!--\; \; -->x$ analytically?

Logarithm problem : Prove that $lo{g}_{3^2}\frac{1}{2}>0$
My approach :
$lo{g}_{3^2}\frac{1}{2}>0$
$\Rightarrow <!--\; \Rightarrow \; -->\frac{1}{2}lo{g}_3\frac{1}{2}>0$
$\Rightarrow <!--\; \Rightarrow \; -->\frac{1}{2}[lo{g}_{3}1-<!--\; -\; -->lo{g}_{3}2]>0$
$\Rightarrow <!--\; \Rightarrow \; -->\frac{1}{2}[0-<!--\; -\; -->lo{g}_{3}2]>0$
$\Rightarrow <!--\; \Rightarrow \; -->-<!--\; -\; -->\frac{1}{2}[lo{g}_{3}2]>0$ { which is false}
Please suggest... thanks...

Logarithmic approximation of $\underset{0}{\overset{N-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}\frac{1}{2i+1}$
Can anyone confirm that it's possible to approximate the sum $\underset{0}{\overset{N-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}\frac{1}{2i+1}$ with the $\frac{\log <!--\; \; -->N}{2}$? And why?
It's clearly visible that the sum has a logarithmic growth over i but it's unclear to me how to prove it.

Taking a logarithmic derivative of a function
I have the following expression:
$\log <!--\; \; -->(1-<!--\; -\; -->\frac{r}{r_s})$
which I would like to take the following derivative of (and where $r_s$ is a constant):
$\frac{d(\log <!--\; \; -->(1-<!--\; -\; -->\frac{r}{r_s}))}{d\log <!--\; \; -->(r)}$
What kind of strategies could I employ to find a solution?

Why $\log <!--\; \; -->(-<!--\; -\; -->2-<!--\; -\; -->y)$ is equivalent to $\log <!--\; \; -->(2+y)$ for restricted values?

Solve the equation $\log <!--\; \; -->(z^2-<!--\; -\; -->1)=i\mathrm{\pi <!--\; \pi \; -->}/2$
I set $z=x+yi$ so:
$\log <!--\; \; -->[(x+yi{)}^2-<!--\; -\; -->1]=\log <!--\; \; -->(x^2+2xyi-<!--\; -\; -->y^2-<!--\; -\; -->1)=\log <!--\; \; -->(r+i\theta )=i\mathrm{\pi <!--\; \pi \; -->}/2$
than I get $x^2-<!--\; -\; -->y^2=1$ and I have no idea how to continue.

Is anyway to prove this: $\underset{k=1}{\overset{n}{\prod <!--\; \prod \; -->}}(a_k)<(1/n^n)\ast <!--\; \ast \; -->(\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}(\sqrt{1+a_k\ast <!--\; \ast \; -->a_{k+1}}){)}^n$
ak and n are positive real number greater than 0.
EDIT: a_{k+1} becomes a_{1} when a_{k}=a_{n}, it is a cylic notation. SORRY.
Any ideas of how to attack the problem?? Thank You.
I don't know if this could help, but the 1/n is also the exponent for the left hand side. I'm thinking maybe of log??
I'm pretty sure that at some point it would be helpful the binominal coefficent?? I don't know.

How is natural log integration broken up into this range? (equation is contained the script)
When I was reading a paper, I found an strange derivation like
$$\begin{gathered} {\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{+\mathrm{\infty <!--\; \infty \; -->}}\mathrm{l}\mathrm{n}(1+e^w)f(w)dw \\ ={\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^0\ln <!--\; \; -->(1+e^w)f(w)+{\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}[\ln <!--\; \; -->(1+e^{-<!--\; -\; -->w})+w]f(w)dw \end{gathered}$$
when $w$ is the normal random variable and $f(w)$ is the normal density.
Why is that natural log integration broken up into that?
I think it just need a simple principle.
Thank you.

How to prove that $\ln <!--\; \; -->(x)<x$ for $x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$
During my calculus homework I need to prove some limits without using L'Hôpital's rule. I have difficulties to show rigorously that $\ln <!--\; \; -->(x)<x$ for big enough x.
For example, I need to find the image of the continues function $f:R\to <!--\; \to \; -->R$ such that for every $x\in <!--\; \in \; -->R$: $|f(x)-<!--\; -\; -->x{e}^{\sqrt{\left|x\right|}}|<x^4$
I've tried to prove that $x{e}^{\sqrt{\left|x\right|}}-<!--\; -\; -->x^4\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$ if $x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$, and then the image of $f$ will be all the reals. Unfortunately, I don't find the way to make it formal enough.

Limit of a harmonic subseries minus "its" logarithm
${\displaystyle \underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{1}{3k-<!--\; -\; -->1}-<!--\; -\; -->\frac{1}{3}\ln <!--\; \; -->(n)}$
I think that inserting the other terms and then subtracting them would not help. I need just the ideea. Thank you.

Why does the method to find out log and cube roots work?
To find cube roots of any number with a simple calculator, the following method was given to us by our teacher, which is accurate to atleast one-tenths.
1)Take the number $X$, whose cube root needs to be found out, and take its square root 13 times (or 10 times) i.e. $\sqrt{\sqrt{\sqrt{\sqrt{....X}}}}$
2)next, subtract 1, divide by 3 (for cube root, and any number n for nth root), add 1.
3) Then square the resultant number (say c) 13times (or 10 times if you had taken out root 10 times) i.e. $c^{2^{2^{....2}}}=c^{2^{13}}$. This yields the answer.
I am not sure whether taking the square root and the squares is limited to 10/13 times, but what I know is this method does yield answers accurate to atleast one-tenths.
For finding the log, the method is similar:-
1)Take 13 times square root of the number, subtract 1, and multiply by $3558$. This yield s the answer.

Why $\int <!--\; \int \; -->{\log }^2<!--\; \; -->(2\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x))dx\ne <!--\; \ne \; -->\frac{{\log }^3<!--\; \; -->(2\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x))\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x)}{3\mathrm{\pi <!--\; \pi \; -->}\cos <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x)}$?
Apparently I completely forgot the basics of calculus after 13 years of not studying it. Why won't:
$\frac{d{\log }^2<!--\; \; -->(2\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x))}{dx}=\frac{{\log }^3<!--\; \; -->(2\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x))\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x)}{3\mathrm{\pi <!--\; \pi \; -->}\cos <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x)}$
UPDATE:
Sorry everybody, I messed up the question big time. Yes, originally it's an integration question why won't:
$\int <!--\; \int \; -->{\log }^2<!--\; \; -->(2\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x))dx=\frac{{\log }^3<!--\; \; -->(2\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x))\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x)}{3\mathrm{\pi <!--\; \pi \; -->}\cos <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}x)}$

Solve some unusual log/exponential equations
I understand about log and exponential equations/functions, but I can't solve these (the numbers are just examples, of course):
$4^x=x+10$
$x^x=3$
$(2x+3x^2{)}^{x+1}=(x-<!--\; -\; -->x^3{)}^{x^2}$
Are there specific algebraic steps that I can follow so that I solve them? Or, if not, what are the other means to find x in each?

Show combination of affine functions and logs has at most one zero
For $x>0$, let
$f(x)=(x+2)\mathsf{l}\mathsf{o}\mathsf{g}(x)-<!--\; -\; -->(x+1)\mathsf{l}\mathsf{o}\mathsf{g}(x+1)$
Can anybody show that the equation $f(x)=0(x>0)$ has at most one solution.

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.

Taylor Series for $\log <!--\; \; -->(x)$
Does anyone know a closed form expression for the Taylor series of the function $f(x)=\log <!--\; \; -->(x)$ where $\log <!--\; \; -->(x)$ denotes the natural logarithm function?

Why can a Complex Logarithm have infinitely many values?
What does this mean, that "Due to the periodicity of the trigonometric functions, a complex logarithm can have infinitely many values"?
$\ln <!--\; \; -->z=\ln <!--\; \; -->(\cos <!--\; \; -->x+i\sin <!--\; \; -->x)=?$