Learn How to Graph Logarithmic Equations

Infinite sum of logarithms
Is there any closed form for this expression
$\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\ln <!--\; \; -->(n+x)$

How to use the logarithm method to solve ${18}^{4x-<!--\; -\; -->3}=(54\sqrt{2}{)}^{3x-<!--\; -\; -->4}$ for $x$?
What value will satisfy this equation:
Please use the logarithm method. I am having a problem in expressing $54\sqrt{2}$ in the power of $18$. My book simply says that it is ${18}^{3/2}$. But I can't understand how.

How do you solve a equation by converting to logarithm form for the problem $3e^{x-<!--\; -\; -->4}+2=83$?
$3e^{x-<!--\; -\; -->4}+2=83$ I understand converting logarithms but i dont understand how to convert e into log form and solve. help would be deeply appreciated.

Why is $\log <!--\; \; -->(n)\in <!--\; \in \; -->o(\frac{n}{\log <!--\; \; -->(n)})$?
This would be equal to:
$\mathrm{\forall <!--\; \forall \; -->}c>0:\mathrm{\exists <!--\; \exists \; -->}{n}_0\in <!--\; \in \; -->\mathbb{N}:\mathrm{\forall <!--\; \forall \; -->}n>n_0:c\log <!--\; \; -->(n)\le \frac{n}{\log <!--\; \; -->(n)}$
For $c=1$ this is obvious, because
$\log <!--\; \; -->(n)\le \sqrt{n}=\frac{n}{\sqrt{n}}\le \frac{n}{\log <!--\; \; -->(n)}$
This estimation doesn't seem to work for $c>1$ though.
Any ideas on how I could prove it for $c>1$?

Is there a property for log(n)/n?
I found a small exercise which I couldn't figure what to do, so I found a solution. Then I tried to understand it and everything went well until I got to this part:
$\frac{1}{8}=\frac{\log <!--\; \; -->(n)}{n}$
Then it just skipped and say that the answer was $n=43$. I was wondering if there is some kind of property for $\log <!--\; \; -->(n)/n$ I don't know about. Or otherwise, how was this solved?
EDIT: This is the exercise Suppose we are comparing implementations of insertion sort and merge sort on the same machine. For inputs of size n, insertion sort runs in 8n2 steps, while merge sort runs in 64n lg n steps. For which values of n does insertion sort beat merge sort?
And this was the solution given:
$8n^2=64n\log <!--\; \; -->(n)$
$n^2=8n\log <!--\; \; -->(n)$
$n=8\log <!--\; \; -->(n)$
$\frac{1}{8}=\frac{log(n)}{n}$

How to solve ${\log }_{x^2-<!--\; -\; -->3}<!--\; \; -->(x^2+6x)<{\log }_x<!--\; \; -->(x+2)$?
How to solve the following inequality
${\log }_{x^2-<!--\; -\; -->3}<!--\; \; -->(x^2+6x)<{\log }_x<!--\; \; -->(x+2)?$?

How to calculate $3\times <!--\; \times \; -->7$ by using logarithm?
This is a story about Newton I read once when I was a child. Now that book is lost and I can only tell you what I remember.
When Newton was young, he had been already famous in curiosity and smart. His family hired a helper. One day, she asked him to go to the market with her because she wasn't good at math. At the market, there was a problem that needed to calculate $3\times <!--\; \times \; -->7$. So the helper asked Newton. After a quick thinking using logarithm, he got the result that $3\times <!--\; \times \; -->7$ must larger than 20 and smaller than 22...
So, my question is how did he do that calculate? How to use logarithm to get the result $20<3\times <!--\; \times \; -->7<22$? Thank you so much.
The story countinues:
... and about to say the result. Before he could finish his math, a near by person had been listened to the conversation and jumped in: "3 times 7 is 21". "Wow, you are smarter than my Newton", said by the helper. "Indeed, you are smarter than Newton", Newton laughed away.

Find all $a$ such that $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\left(\frac{x+a}{x-<!--\; -\; -->a}\right)}^x=e$
Saw this problem and I thought I'd take a shot at it:
Find all $a$ such that
$\underset{x\to <!--\; \to \; --><!--\; \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\left(\frac{x+a}{x-<!--\; -\; -->a}\right)}^x=e.$

Logarithmic Contest Question
The Problem was as follows:
Define $\log \ast <!--\; \ast \; -->(n)$ to be the smallest number of times the log function must be iteratively applied to $n$ to get a result less than or equal to $1$. For example $\log \ast <!--\; \ast \; -->(1000)=2$ since $log(1000)=3$ and $log(3)=0.477\le <!--\; \le \; -->1$. Let $a$ be the smallest integer such that $\log \ast <!--\; \ast \; -->(a)=3$. Computer the number of zeroes in the base 10 representation of $a$
My answer was ${10}^{10}$ but they claimed it to be $9$
My logic was that $\log <!--\; \; -->{10}^{{10}^{10}}={10}^{10}$ and $\log <!--\; \; -->{10}^{10}=10$ and $\log <!--\; \; -->10=1$ and $1\le <!--\; \le \; -->1$ so the $\log \ast <!--\; \ast \; -->{10}^{{10}^{10}}=3$
I presume this means that a smaller answer exists, or I made a logic error somewhere. Can someone show how the smallest answer has 9 zeroes?

Generalize logarithmic coincidences
After playing around with logarithms, I've found the following coincidences:
${\log }_{10}<!--\; \; -->2\approx <!--\; \approx \; -->0.3$, since $2^{10}\approx <!--\; \approx \; -->{10}^3$, and ${\log }_{10}<!--\; \; -->5\approx <!--\; \approx \; -->0.7$, since $5^{1000}\approx <!--\; \approx \; -->{10}^{700}$
I'm sure these are well known. I was just wondering if there was a method or algorithm to generalize these "coincidences" to any base or number. I only have a basic understanding of number theory and would like to learn more.

Proof by induction using logarithms
I have come across a question while studing for my exams prove
${\log }_2<!--\; \; -->x<x\text{ when }x>0$
I know I have to solve it using a base case eg when $x=1$ then assume a inductive step $x=k$ is true but I'm having difficulty trying to solve when $x=K+1$ Can anyone help?

Calculate log of number less than raised to power
I want to calculate the value of 0.9 raised to power 17.I am using the log method.
$17\cdot <!--\; \cdot \; -->\log <!--\; \; -->(0.9)$.Am I doing this correctly?

I wanted to define a similar expression, and search for its properties:
$f(n,x)={}^n\log <!--\; \; -->x=\underset{n\text{ times}}{\underset{⏟<!--\; ⏟\; -->}{\log <!--\; \; -->(\log <!--\; \; -->(\log <!--\; \; -->(\dots <!--\; \dots \; -->\log <!--\; \; -->x)))}},n\in <!--\; \in \; -->{\mathbb{N}}^{+},x\in <!--\; \in \; -->\mathbb{R}$
just looking at how the domain explodes even when $n<8$ baffles me, as I obtained:
${\mathcal{D}}_{f(n\ge <!--\; \ge \; -->2,x)}:x>{}^{n-<!--\; -\; -->2}e\wedge <!--\; \wedge \; -->{\mathcal{D}}_{f(1,x)}:x>0\text{for example, }{\mathcal{D}}_{f(7,x)}:x\gtrsim <!--\; \gtrsim \; -->{10}^{{10}^{{10}^{6.22}}}$
This may be not the only "strange" property of this particular function, but even if I think it can be exploited, I've never found any source (neither textbooks nor online); so in conclusion, is there any application of such function (or functions' class)?

If $\ln <!--\; \; -->x$ is defined via an integral and $e$ defined from $\ln <!--\; \; -->x$, how would you prove that $\ln <!--\; \; -->x$ is the inverse of $e^x$?
This is a somewhat technically specific question about the relationship between $\ln <!--\; \; -->x$ and $e^x$ given one possible definition of $\ln <!--\; \; -->x$
Suppose that you define $\ln <!--\; \; -->x$ as
$\ln <!--\; \; -->x={\int <!--\; \int \; -->}_1^x\frac{dt}{t}$
$\ln <!--\; \; -->x={\int <!--\; \int \; -->}_1^x\frac{dt}{t}$
We can use the connection between the integral definition of $\ln <!--\; \; -->x$ and the harmonic series to show that $\ln <!--\; \; -->x$ grows unboundedly. This function is monotonically increasing, so it should have an inverse mapping from its codomain ($\mathbb{R}$) to its domain (${\mathbb{R}}^{\mathbb{+}}$). Let's call that function ${\ln }^{-<!--\; -\; -->1}<!--\; \; -->x$
Since $\ln <!--\; \; -->x$ grows unboundedly and is zero when $x$ is $1$, we can define $e$ as the unique value such that $\ln <!--\; \; -->e=1$
So here's my question: given these starting assumptions, how would you prove that ${\ln }^{-<!--\; -\; -->1}<!--\; \; -->x=e^x$ (or, equivalently, that $\ln <!--\; \; -->x={\log }_e<!--\; \; -->x$? I'm having a lot of trouble even seeing how you'd get started proving these facts, since basically every calculus fact I know about $e$ and $\ln <!--\; \; -->x$ presumes this result to be true.
Thanks!

How to show $\log <!--\; \; -->a\le <!--\; \le \; -->n(\sqrt[n]{a}-<!--\; -\; -->1)\le <!--\; \le \; -->\sqrt[n]{a}\log <!--\; \; -->a$
Let $b=\sqrt[n]{a}$
How to show: $\log <!--\; \; -->a\le <!--\; \le \; -->n(\sqrt[n]{a}-<!--\; -\; -->1)\le <!--\; \le \; -->\sqrt[n]{a}\log <!--\; \; -->a$?
Thank you ;)

A series converging (or not) to $\ln <!--\; \; -->2$
I have come across the following series, which I suspect converges to $\ln <!--\; \; -->2$
$\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{4^k(2k)}(\genfrac{}{}{0ex}{}{2k}{k}).$
I could not derive this series from some of the standard expressions for $\ln <!--\; \; -->2$. The sum of the first
$100000$ terms agrees with $\ln <!--\; \; -->2$ only up to two digits.
Does the series converge to $\ln <!--\; \; -->2$?

Evaluate $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{1}{\sqrt{x}}{\int <!--\; \int \; -->}_1^x\ln <!--\; \; -->(1+\frac{1}{\sqrt{t}})dt$
If the limit exists with l'Hopital i get
${\displaystyle \underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\ln <!--\; \; -->(1+\frac{1}{\sqrt{x}})}{\frac{1}{2\sqrt{x}}}=2}$
but why can I apply l'Hopital, I mean does the integral ${\displaystyle {\int <!--\; \int \; -->}_1^x\ln <!--\; \; -->(1+\frac{1}{\sqrt{t}})dt}$ really diverge ?
How can I show that ?

Prove symmetry of natural logarithm
Prove that $f(x)=\ln <!--\; \; -->\sqrt{x^2+1}$ is symmetrical in $x=0$
$\ln <!--\; \; -->\sqrt{(x-<!--\; -\; -->a{)}^2+1}=\ln <!--\; \; -->\sqrt{(x+a{)}^2+1}$
$\sqrt{(x-<!--\; -\; -->a{)}^2+1}=\sqrt{(x+a{)}^2+1}$
$(x-<!--\; -\; -->a{)}^2+1=(x+a{)}^2+1$
$x^2-<!--\; -\; -->2ax+a^2+1=x^2+2ax+a^2+1$
$-<!--\; -\; -->2ax=2ax$
$-<!--\; -\; -->2ax=2ax$
$-<!--\; -\; -->x=x$?
I don't know what to do? Is this the proof or did I miss something?

Finding the value of an expression with logarithms
Given that ${\log }_b<!--\; \; -->a=0.74$ and ${\log }_b<!--\; \; -->(a-<!--\; -\; -->1)=0.65$ find the value of the following expression:
${\log }_b<!--\; \; -->(a^4-<!--\; -\; -->1)-<!--\; -\; -->2{\log }_b<!--\; \; -->(a^2+1)+{\log }_b<!--\; \; -->(a^3+a)-<!--\; -\; -->{\log }_b<!--\; \; -->(a+1)$
I tried using log laws to no avail. Help Is much appreciated!

Which is greater: $n^{1.01}$ or $n\cdot <!--\; \cdot \; -->lo{g}_{10}(n)$?
Can someone please explain how the right side can be less than the left side? I have plugged numerous numbers into n and every time I get the left side being less than the right side. My professor is convinced the right side is less than the left side. He has a PHD in math so he should be right. I just don't understand his explanation.
$n^{1.01}<n\cdot <!--\; \cdot \; -->lo{g}_{10}(n)$
${1000}^{1.01}<1000\ast <!--\; \ast \; -->lo{g}_{10}(1000)$
$1071.51<3000$