Learn How to Graph Logarithmic Equations

Logarithmic function with strange bases
Given ${\log }_{4n}<!--\; \; -->40\sqrt{3}={\log }_{3n}<!--\; \; -->45$, find n.
I have rewritten ${\log }_{3n}<!--\; \; -->45$ as ${\displaystyle \frac{{\log }_{4n}<!--\; \; -->45}{{\log }_{4n}<!--\; \; -->3n}}$ and multiplied to get
${\log }_{4n}<!--\; \; -->40\sqrt{3}\cdot <!--\; \cdot \; -->{\log }_{4n}<!--\; \; -->3n={\log }_{4n}<!--\; \; -->45$
but do not know how to continue. Hints would be greatly appreciated, but please don't give me the answer. Thank you.

Calculate ${\int <!--\; \int \; -->}_0^1\frac{{\log }^2<!--\; \; -->x\log <!--\; \; -->(1+x^2)}{1-<!--\; -\; -->x^2}dx$
I found $-<!--\; -\; -->{\displaystyle \frac{{\mathrm{\pi <!--\; \pi \; -->}}^4}{32}}+2G^2+{\displaystyle \frac{7}{4}}\zeta (3)\log <!--\; \; -->2$ where $G$ is the Catalan's constant.

True or false logarithmic branches
Say whether the following are true or false. Give a short proof.
1) $log(-<!--\; -\; -->z)+i\mathrm{\pi <!--\; \pi \; -->}$ is a branch of the logarithmic function whose branch cut is the non-negative real axis
2)If $g(z)$ is a branch of the logarithmic function with domain D and $h(z)$ is a branch analytic in D, then there is an integer $m$ with $h(z)=g(z)+2m\mathrm{\pi <!--\; \pi \; -->}i$
Thanks

Proof ' that $\ln <!--\; \; -->(x)$ converges
Where is the flaw in the following 'proof '?
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\left[\frac{\mathrm{d}}{\mathrm{d}x}\{\ln <!--\; \; -->(x)\}\right]=\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\left[\frac{1}{x}\right]=0⟹<!--\; ⟹\; -->\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}[\ln <!--\; \; -->(x)]=\text{constant}\in <!--\; \in \; -->\mathbb{R},$
therefore $\ln <!--\; \; -->(x)$ converges to some real number. $\mathrm{◻<!--\; ◻\; -->}$

Trouble with Logarithmic Differentiation
Hey guys I'm trying to find the derivative of this equation using logarithmic differentiation but I'm having some trouble. Wolfram Alpha is giving me different answers and I'm having difficulty understanding the steps it spits out. Could somebody take a look at what I'm doing and tell me where I went wrong?
The problem I need to differentiate is:
$y=(\sin <!--\; \; -->x{)}^{\ln <!--\; \; -->x}$
My first step is to take the logarithm of both sides and factor out the exponent.
$\ln <!--\; \; -->(y)=\ln <!--\; \; -->x\ast <!--\; \ast \; -->\ln <!--\; \; -->(\sin <!--\; \; -->x)$
Next I try to differentiate both sides, using the rules for logarithms:
$\frac{1}{y}\ast <!--\; \ast \; -->y^{\prime }=\frac{1}{x}\frac{1}{\sin <!--\; \; -->x}\cos <!--\; \; -->x\Rightarrow <!--\; \Rightarrow \; -->\frac{1}{y}\ast <!--\; \ast \; -->y^{\prime }=\frac{\cos <!--\; \; -->x}{x\sin <!--\; \; -->x}$
Which finally renders:
$\frac{dy}{dx}=\frac{\cos <!--\; \; -->x}{x\sin <!--\; \; -->x}(\sin <!--\; \; -->x{)}^{\ln <!--\; \; -->x}$
But this doesn't seem to be the right answer, could somebody help me out?

Number raised to log expression
I am struggling with what I think should be some a basic log problem:
Show that $3^{lo{g}_{2}n}=n^{lo{g}_{2}3}$
I know that $3^{lo{g}_{3}n}=n$ and $lo{g}_{2}n=lo{g}_{3}n/lo{g}_{3}2$
I was attempting something similar to:
$3^{lo{g}_{3}n/lo{g}_{3}2}=3^{lo{g}_{3}n-<!--\; -\; -->lo{g}_{3}2}$ but then I got stuck. Am I on the right track by using the change of base and then subtracting?
EDIT: I'm not trying to exactly 'show that' the two expressions are equal. I am looking to figure out how to simplify the expression on the left to the one on the right

How to solve exponential inequality with $x$
I need to solve the following inequality.
$\ln <!--\; \; -->(x)-<!--\; -\; -->x>0.$
I oddly remember that it can only be done by using the graph... Is it true?
I have the same problem with
$e^x(x-<!--\; -\; -->1)>-<!--\; -\; -->2.$
Thanks!

Solve ${\log }_2<!--\; \; -->(1+\frac{1}{x-<!--\; -\; -->1})<1$
I don't get how my teacher got two different equations out of the one. One is $>0$ and the other one is $<2$. Be detailed please.

Natural logarithm problem
I'm kind of confused on how to solve this problem! Any guidance/advice would be appreciated. Thanks!
$e^{-9}{e}^{-2}{e}^9$

Evaluate the integral. $\int <!--\; \int \; -->x^2\log <!--\; \; -->(4x)dx$
The problem is $\int <!--\; \int \; -->x^2\log <!--\; \; -->(4x)dx$
Here $\ln$ refers to the natural logarithm.
So far, I know $u=x^2$ and $du=2x(dx)$
So $dv=\ln <!--\; \; -->(4x)dx$ and $v=1/x$, but I don't know where to go from here.

Determine all positive numbers $a$ for which the curve $y=a^x$ intersects the line $y=x$ without calculus
The answer is $0<a<e^{1/e}$ , but how to find it? Is it a system of equations? Which ones? I just need an idea at least, because I'm stuck. If it is impossible without calculus, solve it with it.

problem solving logarithmic equation and reaching an equivalence
ok so i've had a problem trying to simplify the $\ln <!--\; \; -->[\sqrt{1+\frac{u^2}{a^2}}+\frac{u}{a}]$ and this is supposed to be equal to : $\ln <!--\; \; -->[\sqrt{a^2+u^2}+u]$
how is this posible ?? i've tried to solve this for more than 2 hours and couldn't get to this equivalence. any suggestions ?

Solution to $x=100e^{-<!--\; -\; -->x/100}$?
How do you solve
$x=100e^{-<!--\; -\; -->x/100}$
If I use $\ln$ then $\ln <!--\; \; -->(x/100)=-<!--\; -\; -->x/100$. How do I get from that to $x=56.7$?

How to solve this logarithmic equation whose expressions have different bases?
I have been trying to solve the following equation for a while and i can't seem to figure it out, your help would be greatly appreciated. Here is the equation: $3^x$

Expansion of Logarithms with Cube Roots
Does the following expand to the following
${\log }_6<!--\; \; -->({11}^6\sqrt[3]{12})$
= $6{\log }_6<!--\; \; -->(11)+{\log }_6<!--\; \; -->(\sqrt[3]{12})$

Stuck on derivative of logarithm of sum of exponentials
let's say that I need to calculate the following expression:
$\frac{\mathrm{\partial <!--\; \partial \; -->}\mathrm{l}\mathrm{o}\mathrm{g}(\mathrm{e}\mathrm{x}\mathrm{p}(w_1\ast <!--\; \ast \; -->x_1+b_1)+\mathrm{e}\mathrm{x}\mathrm{p}(w_2\ast <!--\; \ast \; -->x_2+b_2))}{\mathrm{\partial <!--\; \partial \; -->}{w}_1}$
How do I start? The rules that I know about derivation, exponential and logarithm are not very useful in this situation.

what's the relationship with log(sum) and sum(log)?
hi I'm a little confused about the log(sum) function and sum(log) function. In special, what's the relationship between these two terms?
$-<!--\; -\; -->\log <!--\; \; -->\underset{i}{\sum <!--\; \sum \; -->}{a}_i\underset{i}{\sum <!--\; \sum \; -->}{b}_i$
$-<!--\; -\; -->\underset{i}{\sum <!--\; \sum \; -->}\log <!--\; \; -->(a_i+b_i)$
given a negative log-likelihood of an observation set:
$\mathbf{L}=-<!--\; -\; -->\underset{i,j}{\sum <!--\; \sum \; -->}\log <!--\; \; -->({\mathrm{\pi <!--\; \pi \; -->}}_a{M}_{i,j}+{\mathrm{\pi <!--\; \pi \; -->}}_b{N}_{i,j})$
where C is the constant parameter. ${\mathrm{\pi <!--\; \pi \; -->}}_a$+${\mathrm{\pi <!--\; \pi \; -->}}_b$=1 are proportion of the two component, given the instance $O_{ij}$
Lemma1
$$\begin{gathered} -<!--\; -\; -->\log <!--\; \; -->\underset{k=1}{\overset{K}{\sum <!--\; \sum \; -->}}{f}_k(x)=\underset{\mathrm{\Phi <!--\; \Phi \; -->}(x)\in <!--\; \in \; -->{\mathrm{\Delta <!--\; \Delta \; -->}}_{+}}{min}\underset{k=1}{\overset{K}{\sum <!--\; \sum \; -->}}\{{\mathrm{\Phi <!--\; \Phi \; -->}}_k(x)[\log <!--\; \; -->{\mathrm{\Phi <!--\; \Phi \; -->}}_k(x)-<!--\; -\; -->log(f_k(x)]\} \\ s.t.\sum <!--\; \sum \; -->{\mathrm{\Phi <!--\; \Phi \; -->}}_k(x)=1,{\mathrm{\Phi <!--\; \Phi \; -->}}_k(x)\in <!--\; \in \; -->(0,1) \end{gathered}$$
proof
$$\begin{gathered} RHS=\underset{k=1}{\overset{K}{\sum <!--\; \sum \; -->}}{\mathrm{\Phi <!--\; \Phi \; -->}}_k(x)\log <!--\; \; -->\frac{{\mathrm{\Phi <!--\; \Phi \; -->}}_k(x)}{f_k(x)} \\ >=\underset{k=1}{\overset{K}{\sum <!--\; \sum \; -->}}{\mathrm{\Phi <!--\; \Phi \; -->}}_k(x)\log <!--\; \; -->\frac{\underset{k=1}{\overset{K}{\sum <!--\; \sum \; -->}}{\mathrm{\Phi <!--\; \Phi \; -->}}_k(x)}{\underset{k=1}{\overset{K}{\sum <!--\; \sum \; -->}}{f}_k(x)}(log-<!--\; -\; -->suminequality) \\ =-<!--\; -\; -->\log <!--\; \; -->\underset{k=1}{\overset{K}{\sum <!--\; \sum \; -->}}{f}_k(x)(\sum <!--\; \sum \; -->{\mathrm{\Phi <!--\; \Phi \; -->}}_k(x)=1) \end{gathered}$$
Let:
$\mathit{C}=\underset{i,j}{\sum <!--\; \sum \; -->}{\mathrm{\Phi <!--\; \Phi \; -->}}_a^{i,j}(\log <!--\; \; -->{\mathrm{\Phi <!--\; \Phi \; -->}}_a^{i,j}-<!--\; -\; -->\log <!--\; \; -->({\mathrm{\pi <!--\; \pi \; -->}}_a{M}_{i,j}))+{\mathrm{\Phi <!--\; \Phi \; -->}}_b^{i,j}(\log <!--\; \; -->{\mathrm{\Phi <!--\; \Phi \; -->}}_B^{i,j}-<!--\; -\; -->\log <!--\; \; -->({\mathrm{\pi <!--\; \pi \; -->}}_b{N}_{i,j}))$
given the constraint, that for each $(i,j)$, ${\mathrm{\Phi <!--\; \Phi \; -->}}_a^{i,j}+{\mathrm{\Phi <!--\; \Phi \; -->}}_b^{i,j}=1$
Then how to prove:
Minimize $C$ equals minimize $L$?
we have
$minC=-<!--\; -\; -->\log <!--\; \; -->\sum <!--\; \sum \; -->({\mathrm{\pi <!--\; \pi \; -->}}_a{M}_{i,j})-<!--\; -\; -->\log <!--\; \; -->\sum <!--\; \sum \; -->({\mathrm{\pi <!--\; \pi \; -->}}_b{N}_{i,j})$
then the next step is how to prove the relationship between $minC$ and $L$?

Inverse of the function $-<!--\; -\; -->\log <!--\; \; -->(1-<!--\; -\; -->[1-<!--\; -\; -->e^{-<!--\; -\; -->x^{\mathrm{\alpha <!--\; \alpha \; -->}}}{]}^{\mathrm{\beta <!--\; \beta \; -->}})$
I have a function as follows, I would like to get the inverse of this function. What is the inverse of $f(x)$?
$y=f(x)=-<!--\; -\; -->\log <!--\; \; -->(1-<!--\; -\; -->[1-<!--\; -\; -->e^{-<!--\; -\; -->x^{\mathrm{\alpha <!--\; \alpha \; -->}}}{]}^{\mathrm{\beta <!--\; \beta \; -->}})$
Is my answer correct?
$f^{-<!--\; -\; -->1}(x)=(-<!--\; -\; -->\log <!--\; \; -->(1-<!--\; -\; -->(1-<!--\; -\; -->{\exp }^{-<!--\; -\; -->x}{)}^{1/\mathrm{\beta <!--\; \beta \; -->}}){)}^{1/\mathrm{\alpha <!--\; \alpha \; -->}}$

Obtain the set of real numbers $c$
Show that there exists a positive real number $x\ne <!--\; \ne \; -->2$ such that ${\log }_2<!--\; \; -->x=\frac{x}{2}$ . Hence obtain the set of real numbers $c$ such that $\frac{{\log }_2<!--\; \; -->x}{x}$$=c$ has only one real solution.

Can't find solution to Calculus 8th (Adams, Essex) limit problem
I've been sitting here for hours trying to find a solution to his problem:
If you have the function $g(y)$, which is the inverse of
$$\begin{gathered} f(x)=x^x \\ {e}^{-<!--\; -\; -->1}\le <!--\; \le \; -->x<\mathrm{\infty <!--\; \infty \; -->}, \end{gathered}$$
Show that
$\underset{y\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\displaystyle \frac{g(y)\ln <!--\; \; -->(\ln <!--\; \; -->y)}{\ln <!--\; \; -->y}}=1$
Hint: Start with the equation $y=x^x$ and take the ln of both sides twice.