Learn How to Graph Logarithmic Equations

Steps to Graph Exponential Equations & Absolute Value
how to sketch: $-<!--\; -\; -->e^{|-<!--\; -\; -->x-<!--\; -\; -->1|}+2$
Can someone clarify:
$|f(x)|:$ we draw $f(x)$ and then reflect the ($-<!--\; -\; -->y$ parts) in the $x$-axis
$f|(x)|:$ we draw $f(x)$ and then reflect the ($-<!--\; -\; -->x$ parts) in the $y$-axis (symmetry on left and right hand side), can someone correct me here!
The Steps to sketch the above equation:
1.The original equation is e^x which then becomes e^|x| which shows to have undergone f|(x)|
2.We sketch -e^(-x-1) without applying the absolute value
3.Then we apply the absolute value, by reflecting along the turning point
4.Then we shift the graph up 2 units
so basically:
1.draw the e^x graph
2.apply the reflections/dilations/horizontal translations
3.apply f(|x|)
4.apply the vertical transformation
can someone correct me here.

Having trouble solving $\log <!--\; \; -->(x-21)=2-\log <!--\; \; -->x$ for $x$
I'm having trouble with this problem:
$\log <!--\; \; -->(x-21)=2-\log <!--\; \; -->x$, solve for $x$
I'm coming up with $x=-<!--\; -\; -->5$ but that can't be right.

When does $\sum <!--\; \sum \; -->\frac{1}{(n\ln <!--\; \; -->n{)}^a}$ converge?
For $a>1$, I used the limit comparison test as ${\displaystyle \frac{\frac{1}{(n\ln <!--\; \; -->n{)}^a}}{\frac{1}{n^a}}\to <!--\; \to \; -->0}$ as $n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$ so that ${\displaystyle \sum <!--\; \sum \; -->\frac{1}{(n\ln <!--\; \; -->n{)}^a}}$ converges together with ${\displaystyle \sum <!--\; \sum \; -->\frac{1}{n^a}}$ and for $a\le <!--\; \le \; -->\frac{1}{2}$ notice that ${\displaystyle \sum <!--\; \sum \; -->\frac{1}{(n\ln <!--\; \; -->n{)}^a}>\sum <!--\; \sum \; -->\frac{1}{n^{2a}}}$ and since RHS diverges we can see ${\displaystyle \sum <!--\; \sum \; -->\frac{1}{(n\ln <!--\; \; -->n{)}^a}}$ diverges . But what happens if $a\in <!--\; \in \; -->(\frac{1}{2},1)$? Or is there any other method to cover these cases on one step?

Two different solutions of the same integral
Considering
$\int <!--\; \int \; -->\frac{\ln <!--\; \; -->(x+1)}{2(x+1)}dx$
I first solved it seeing it similar to the derivative of ${\ln }^2<!--\; \; -->(x+1)$ so multiplying by $\frac{2}{2}$ the solution is
$\int <!--\; \int \; -->\frac{\ln <!--\; \; -->(x+1)}{2(x+1)}dx=\frac{{\ln }^2<!--\; \; -->(x+1)}{4}+const.$
. But then we can solve it using by parts' method and so this is the solution that I found:
$\frac{1}{2}\int <!--\; \int \; -->\frac{\ln <!--\; \; -->(x+1)}{(x+1)}dx=\frac{1}{2}\ln <!--\; \; -->(x+1)\ln <!--\; \; -->(x+1)-<!--\; -\; -->\frac{1}{2}\int <!--\; \int \; -->\frac{\ln <!--\; \; -->(x+1)}{x+1}dx$
Seeing it as an equation I brought the integral $-<!--\; -\; -->\frac{1}{2}\int <!--\; \int \; -->\frac{\ln <!--\; \; -->(x+1)}{x+1}dx$ to the left so that I obtain
$\int <!--\; \int \; -->\frac{\ln <!--\; \; -->(x+1)}{(x+1)}dx=\frac{1}{2}\ln <!--\; \; -->(x+1)\ln <!--\; \; -->(x+1)+const.$
so
$\int <!--\; \int \; -->\frac{\ln <!--\; \; -->(x+1)}{(x+1)}dx=\frac{1}{2}{\ln }^2<!--\; \; -->(x+1)+const.$
. I know that the first solution is correct but I used to way of solution that seem to be correct. How is it possible? Where is the mistake? Thank you in advance for your help!

Exponential continuous growth $\ln <!--\; \; -->a$ vs. $r$? Huh?
So, given a simple population continuous growth problem, it seems that the entirety of the internet uses $P=P_0{e}^{rt}$ where $P$ is the population over time, $P_0$ is the initial population, $r$ is the percentage of change, and $t$ is time.
But this class is asserting that $P=P_0{e}^{(\ln <!--\; \; -->a)(t)}$ where $a=1+r$, and everything else stays the same.
So lets say a population starts at $10,000$ with $10$ continuous growth rate. What is the population in $10$ years?
The internet says that $P=10,000e^{(.10)(10)}=27,182$
But the class says that $P=10,000e^{(\ln <!--\; \; -->1.10)10}=25,937$ ($a=1+r$ which is $1.10$)
but that equals the usual constant growth of $P=P_0{a}^t=25937$
What the heck? What am I missing?

Convergence $\underset{n=2}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{\sqrt[n]{n^p}}{n\ln <!--\; \; -->n}$
Help please. I need to check the convergence
$\underset{n=2}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{\sqrt[n]{n^p}}{n\ln <!--\; \; -->n}$
Tried with Leibniz, but can't check monotony.

$\underset{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{2^k}{\mathrm{\gamma <!--\; \gamma \; -->}}\log <!--\; \; -->\mathbb{E}[e^{-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}\frac{X}{2^k}}]$
I am trying to find a limit for this expression
$\underset{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{2^k}{\mathrm{\gamma <!--\; \gamma \; -->}}\log <!--\; \; -->\mathbb{E}[e^{-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}\frac{X}{2^k}}]$
I have so far found these bounds:
$\frac{1}{\mathrm{\gamma <!--\; \gamma \; -->}}\log <!--\; \; -->{(\mathbb{E}[e^X])}^{-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}}\le <!--\; \le \; -->\underset{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{2^k}{\mathrm{\gamma <!--\; \gamma \; -->}}\log <!--\; \; -->\mathbb{E}[e^{-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}\frac{X}{2^k}}]\le <!--\; \le \; -->\frac{1}{\mathrm{\gamma <!--\; \gamma \; -->}}\log <!--\; \; -->\mathbb{E}[e^{-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}X}]$
I was wondering if there are better bounds in particular a better lower bound estimate. $\mathrm{\gamma <!--\; \gamma \; -->}\in <!--\; \in \; -->(0,\mathrm{\infty <!--\; \infty \; -->})$ and $X\in <!--\; \in \; -->L^{\mathrm{\infty <!--\; \infty \; -->}}(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{F},\mathbb{P})$

Exponential function to logarithmic function
i'm stuck on completing this equations. Is this correct?
$z=a{e}^{-<!--\; -\; -->bt}$
$\ln <!--\; \; -->(z)=\ln <!--\; \; -->(a)+\ln <!--\; \; -->(e^{-<!--\; -\; -->bt})$
$\ln <!--\; \; -->(z)=\ln <!--\; \; -->(a)+(1)(-<!--\; -\; -->bt)$
$\ln <!--\; \; -->(z)=\ln <!--\; \; -->(a)-<!--\; -\; -->bt$

Logarithm multiplication property error, can't figure out why.
I know there is a mistake and where it is but I can't figure out why.
Equation:
$3+2({12}^{x+1})=291$
From here I do:
$$\begin{gathered} 2({12}^{x+1})=291-<!--\; -\; -->3 \\ 2({12}^{x+1})=288 \end{gathered}$$
Then I take the natural logarithms on both sides;
$$\begin{gathered} \ln <!--\; \; -->(2\ast <!--\; \ast \; -->{12}^{x+1})=\ln <!--\; \; -->(288) \\ \ln <!--\; \; -->(2\ast <!--\; \ast \; -->{12}^{x+1})=\ln <!--\; \; -->(2\ast <!--\; \ast \; -->{12}^2) \end{gathered}$$
Now I apply multiplication property so $\ln <!--\; \; -->(a\ast <!--\; \ast \; -->b)$ should equal $\ln <!--\; \; -->(a)+\ln <!--\; \; -->(b)$ But here it seems that I'm making a mistake, can't figure out why:
$\ln <!--\; \; -->(2)+\ln <!--\; \; -->({12}^{x+1})=\ln <!--\; \; -->(2)+\ln <!--\; \; -->({12}^2)$
subtract $\ln <!--\; \; -->(2)$ on both sides:
$\ln <!--\; \; -->({12}^{x+1})=\ln <!--\; \; -->({12}^2)$
Then it should be:
$$\begin{gathered} (x+1)\ln <!--\; \; -->(12)=2\ln <!--\; \; -->(12) \\ x+1=2 \\ x=1 \end{gathered}$$
but it's not the right result..
Thanks in advance :)

Find the value of the Logarithmic Expression
Why is ${\log }_6<!--\; \; -->1$ equal to $0$?

Differential equation $x{y}^{\prime }+2y=0$ and the form of arbitrary constant in its general solution
If I'm solving the differential equation in the title I will get to:
$\log <!--\; \; -->(y)=-<!--\; -\; -->2\log <!--\; \; -->(x)+c$
then I'll get $y=e^c/x^2$ eith arbitrary constant c.
So I know I can write $y=d/x^2$ where d is an arbitrary positive constant. But now it's easy to see that d can be negative or zero as well. How can I see that from the derivation? Can I reason from the derivation that d can be arbitrary? (I know I can just show d can be arbitrary e.g. by calculating y′ and see that d cancels out.)

How to solve if I have ln on both sides of equation?
I thought this would be a common problem but googling hasn't helped.
If I have $\ln <!--\; \; -->(ex)=\ln <!--\; \; -->(y)$ what the next step to solve for $y$?

How do I find the inverse of this exponential function?
$x=-<!--\; -\; -->3(3^{-<!--\; -\; -->x})+9$
I know the steps up until a certain point.
$x=-<!--\; -\; -->3(3^{-<!--\; -\; -->y})+9$
$x-<!--\; -\; -->9=-<!--\; -\; -->3(3^{-<!--\; -\; -->y})$
$\frac{(x-<!--\; -\; -->9)}{-<!--\; -\; -->3}=3^y$
$ln(\frac{x-<!--\; -\; -->9}{-<!--\; -\; -->3})=-<!--\; -\; -->y\ast <!--\; \ast \; -->ln3$
Not sure what to do from here. I know I have to get y by itself but thats it. Can anyone help please?

Multivariable-calculus, logarithms
I got the function $f(x,y)=\ln <!--\; \; -->(1+x^2+y^2)$. There are three tasks to answer.
a)Decide the function´s stationary points and classify them if possible. Here I got the answer to $(0,0)$ is a local maximum point.
b)Decide a Here I got the answer $(0,0)$
c)Now limit the domain of definition to $x^2+y^2\le 1$. Decide the function´s biggest and lowest value and the range.
Please help me with task c). The answer should be: lowest value: $f(0,0)=0$. Biggest value: $f(a,b)=\ln <!--\; \; -->(2)$ for all $a^2+b^2=1$. The range is $0\le z\le \ln <!--\; \; -->(2)$. How do I get there?

Struggling with a Form of Logarithm question during my revision
I am doing AS Mathematics In the UK under the examining board edexcel.
I came across this question in a List of exam questions given to me by my teacher However I can't work out how to do it.
$2^x{4}^{x+1}=8$
Now I understand that $4^{x+1}$ is the same as $(2^2{)}^x$ therefore the $\log$ is base $2$
So $\log$ base $2$ of $8$ = 3
And then I'm stuck. Can anyone help me on this? It'd be greatly appreciated :D

How to solve this logarithm equation
${\log }_2<!--\; \; -->\{{\log }_3<!--\; \; -->[{\log }_4<!--\; \; -->\left(x^{3x}\right)]\}=0$
How would I go about solving this? I tried doing ${\log }_4<!--\; \; -->(x^{3x}))=0$ but I don't know how to incorporate the other logs

Find the minimum value of $S$
Let $a,b,c$ be real numbers greater than 1.
Let $S={\log }_a<!--\; \; -->bc+{\log }_b<!--\; \; -->ac+{\log }_c<!--\; \; -->ba$
Then find the minimum value of $S$

Intuitive understanding of logarithms
I know logarithms are supposed to be the inverse of exponential functions, and while this makes sense, it seems to me that a more intuitive and significant property is
$\log <!--\; \; -->(ab)=\log <!--\; \; -->(a)+\log <!--\; \; -->(b)$
So in this way, the logarithm is a fundamental relationship between addition and multiplication. Should logarithms in schools be taught this way? Should I think of them primarily in this way?
EDIT: This probably related to the fact that the only continuous functions $f$ that satisfy $f(x+y)=f(x)f(y)$ are exponential functions (there are apparently some super-weird non-continuous non-exponential functions that satisfy that multiplicativity but I have no idea what they are).

Evaluate ${\int <!--\; \int \; -->}_0^1{\log }^2<!--\; \; -->(1-<!--\; -\; -->x){\log }^2<!--\; \; -->(x)dx$
I have no idea where to even start. WolframAlpha cant compute it either.
${\int <!--\; \int \; -->}_0^1{\log }^2<!--\; \; -->(1-<!--\; -\; -->x){\log }^2<!--\; \; -->(x)dx$
I think it can be done with series, but I am not sure, can someone help a little so I can get a start??
Thanks!

Algebraically, how are $-<!--\; -\; -->\ln <!--\; \; -->|\csc <!--\; \; -->x+\cot <!--\; \; -->x|+C$ and $\ln <!--\; \; -->|\csc <!--\; \; -->x-<!--\; -\; -->\cot <!--\; \; -->x|+C$ equal?
I know both of these are the answer to $\int <!--\; \int \; -->\csc <!--\; \; -->xdx$, and I am able to work them out with calculus using the formulas:
$\int <!--\; \int \; -->\csc <!--\; \; -->xdx$
$=\int <!--\; \int \; -->\csc <!--\; \; -->x\frac{\csc <!--\; \; -->x-<!--\; -\; -->\cot <!--\; \; -->x}{\csc <!--\; \; -->x-<!--\; -\; -->\cot <!--\; \; -->x}dx$
and:
$=\int <!--\; \int \; -->\csc <!--\; \; -->x\frac{\csc <!--\; \; -->x+\cot <!--\; \; -->x}{\csc <!--\; \; -->x+\cot <!--\; \; -->x}dx$
Still, when looking at the results, $-<!--\; -\; -->\ln <!--\; \; -->|\csc <!--\; \; -->x+\cot <!--\; \; -->x|+C$ and $\ln <!--\; \; -->|\csc <!--\; \; -->x-<!--\; -\; -->\cot <!--\; \; -->x|+C$ , I don't see how these are algebraically equivalent. Perhaps I'm just unaware of some algebra rule (that is likely!). I tried using the Laws of Logs and that doesn't help. Or maybe I'm missing some trig trick.