Learn How to Graph Logarithmic Equations

Evaluate:
${12}^{\log }12(3)$

Logarithm problems with different bases
1.${\log }_a<!--\; \; -->b\times <!--\; \times \; -->{\log }_b<!--\; \; -->a=$?
2.${\log }_a<!--\; \; -->b+{\log }_b<!--\; \; -->a=\sqrt{29}$
What is ${\log }_a<!--\; \; -->b-<!--\; -\; -->{\log }_b<!--\; \; -->a=$?
3.
What is b in the following:
${\log }_b<!--\; \; -->3+{\log }_b<!--\; \; -->11+{\log }_b<!--\; \; -->61=1$
and
4.
$\frac{1}{lo{g}_{2}x}+\frac{1}{lo{g}_{25}x}-<!--\; -\; -->\frac{3}{{\log }_8<!--\; \; -->x}=\frac{1}{{\log }_b<!--\; \; -->x}$
What is b?
Can anyone help me solve these?

Find the range of values that $x$ can take if $9{\log }_x<!--\; \; -->5={\log }_5<!--\; \; -->x$
I'm stuck on a homework question about logarithms. I can't work out how to do it, and all I've managed to do is turn $9{\log }_x<!--\; \; -->5$ into ${\log }_x<!--\; \; -->5^9$. Can anyone guide me onto the right path to solve this?

What is the meaning of this Wolfram Alpha result when calculating $3^p=4^q$?
I would like to know are the some $p\in <!--\; \in \; -->\mathbb{N}$ and $q\in <!--\; \in \; -->\mathbb{N}$ for $3^p=4^q$ except the trivial $p=q=0$. So, I entered the expression into Wolfram Alpha, which returned the result for $q$:
${\displaystyle q=\frac{\log <!--\; \; -->(3^p)}{\log <!--\; \; -->(4)}+\frac{2\mathrm{\pi <!--\; \pi \; -->}i{c}_1}{\log <!--\; \; -->(4)}}$
Nothing is said about what is $c_1$. Arbitrary constant? If $c_1\ne <!--\; \ne \; -->0$ the result could be complex-valued I guess. I tried some calculations using the formula but the results do not make sense.

Integration by parts: $\int <!--\; \int \; -->x\ln <!--\; \; -->x^{2}dx$
Problem: $\int <!--\; \int \; -->x\ln <!--\; \; -->x^{2}dx$
So what I did first was make $u=\ln <!--\; \; -->x^2$ and $dv=x$
Then I solved by getting the derivative of $u$ and the anti derivative of $dv$ and I got $du=1/x^2$ and $v=x^2/2$ then I did the formula
$\int <!--\; \int \; -->udv=uv-<!--\; -\; -->\int <!--\; \int \; -->vdu$
which then after plugging in the numbers and simplifying got me
$\frac{x^2}{2}\ln <!--\; \; -->x^2-<!--\; -\; -->\frac{1}{2x}+C$
Is this the right way to do the problem and answer?

How to generate $\log$ function that intersects at $(0,1)$ and $(1,0)$?
I apologize for any incorrect or missing formatting, first time posting in the math stack exchange. It's been a few years since I've done any kind of calculus, so I remember nothing at all, which is probably the reason why I find my self stumped so early in my calculations.
I'm looking to generate a logarithmic algorithm that will follow a plot that will pass the following points: $(0,1)$ and $(1,0)$. My end goal is to generate a programming function that will return a very simple repulsion force ceofficient value based on a distance between two points. I would like, however, that the force drop (logarithmic?) the further these two points are, and grow (exponentially?) the closer they get. This is in no way an accurate calculation I'm looking for, but merely a non-linear function to return me a scalar coefficient between 0 and 1 that I can manipulate later on.
I want the function to cross at $(1,0)$ and $(0,1)$ so I get a coefficient between $0$ and $1$. In my function, any Y value lower than $0$ will be floored at $0$ and likewise for $Y$ values over $1$ (where $x<0$, which is possible since my $X$ value in my algorithm will likely be an offsetted value of the distance between two points).
That being said, with the following few assumptions:
1.$y=0$ when $x=1$
2.$y=1$ when $x=0$
3.$y<0$ when $x>1$
4.$y>1$ when $x<0$
5.$y=Alog(x+B)+C$
6.assuming log is log based 10
I tried to deduce the $A,B,C$ constants in order to generate a formula.
Here is what I've done so far:
1.From assumption #1, I can generate $1=Alog(B)+C$
2.From assumption #2, I can generate $0=Alog(1+B)+C$
3.I make both equations equals by transforming equation from 1. to $0=Alog(B)+C-<!--\; -\; -->1$
4.Making the following equality: $Alog(B)+C-<!--\; -\; -->1=Alog(1+B)+C$
5.I can scratch both $C$ constants out (is it safe to assume $C=0$, or $C$ can be equal to any arbitrary value, without changing the equation's plot?): $Alog(B)-<!--\; -\; -->1=Alog(1+B)$
6.Dividing each part by A gets me: ${\displaystyle \frac{Alog(B)}{A}}-<!--\; -\; -->{\displaystyle \frac{1}{A}}={\displaystyle \frac{Alog(1+B)}{A}}$
7.Allowing me to reduce both logs: $\log <!--\; \; -->(B)-<!--\; -\; -->{\displaystyle \frac{1}{A}}=log(1+B)$
8.If I put both logs on the same side: $log(B)-<!--\; -\; -->log(1+B)={\displaystyle \frac{1}{A}}$
9.I can combine them into: $\log <!--\; \; -->({\displaystyle \frac{B}{1+B}})={\displaystyle \frac{1}{A}}$
And I'd say this is where I'm stuck; I have two unknowns ($A$ and $B$) with one equation to solve. I don't know where to go from here. What's the next step?

Find the inverse of the function
Find the inverse of the function $f(x)=-<!--\; -\; -->2\cdot <!--\; \cdot \; -->4^{2(x-<!--\; -\; -->3)}-<!--\; -\; -->1$

Prove that $\ln <!--\; \; -->x\le <!--\; \le \; -->x-<!--\; -\; -->1$
I need help with this proof for my real analysis class. it is on the practice sheets and we do NOT get an answer. I proved $\ln <!--\; \; -->(x)<x-1$ for all $x>1$ by contradiction but cannot do this one.
Prove that $\ln <!--\; \; -->(x)\le <!--\; \le \; -->x-1$ for all $x>0$
i believe you need to use MVT, I cannot use the famous inequality $e^x>x+1$ for all $x>0$

Trouble evaluating the sum involving logarithm
I was trying to solve this problem: Closed form for ${\int <!--\; \int \; -->}_0^1\log <!--\; \; -->\log <!--\; \; -->(\frac{1}{x}+\sqrt{\frac{1}{x^2}-<!--\; -\; -->1})\mathrm{d}x$
In the procedure I followed, I came across the following sum:
$\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^{k-<!--\; -\; -->1}k(\frac{\ln <!--\; \; -->(2k+1)}{2k+1}-<!--\; -\; -->\frac{\ln <!--\; \; -->(2k-<!--\; -\; -->1)}{2k-<!--\; -\; -->1})$
I cannot think of any approaches which would help me in evaluating the sum.
Any help is appreciated. Thanks!

Solve logarithmic equation: $2{\log }_7<!--\; \; -->(x+2)-<!--\; -\; -->{\log }_7<!--\; \; -->(3x+10)=0$
Please, can someone check if this is the right answer
$x=-<!--\; -\; -->2\pm <!--\; \pm \; -->\sqrt{3x+10}$
Thank you.

Help me to solve math homework on logarithmic
How to solve this math home work? Please help..
What is the value of $\log <!--\; \; -->\left({\displaystyle \frac{i\mathrm{\pi <!--\; \pi \; -->}}{2}}\right)$?
I got to know the answer is "${\displaystyle \frac{i\mathrm{\pi <!--\; \pi \; -->}}{2}}$", but don't know how to solve it. Please help me.
Thanks a lot in advance.

How to calculate arithmetic mean of log values
I am working with really small values of probabilities and that is why their log values are used. So for example, let probA and probB be some normal values of probabilities of two events and because they are very small, I use log values of them, $ln(probA)$ and $ln(probB)$
I would like to compute arithmetic mean of this two values to compare them with other such pairs. Because they are very small, it is not a good idea to compute arithmetic mean like
$\frac{e^{ln(probA)}+e^{ln(probB)}}{2}$
Would it be mathematically correct only to compute arithemtic mean of the log values? Or is there any other better way to do it?

Show $(\ln <!--\; \; -->(x^2){)}^2-<!--\; -\; -->(\ln <!--\; \; -->x{)}^2=3(\ln <!--\; \; -->x{)}^2$
I read an example on integrals.
I can't see how
$(\ln <!--\; \; -->(x^2){)}^2-<!--\; -\; -->(\ln <!--\; \; -->x{)}^2=3(\ln <!--\; \; -->x{)}^2.$

Find the largest possible root of a number that is whole
Using 8 as an example radicand, the degree would be 3 because ∜8 is not a whole number, while √8 is not the largest possible whole root. This type of problem is easy to calculate mentally with small numbers, but for large numbers it gets tricky. Is there a way to calculate it without iterating through degrees until the right one is found? It seems the answers may involve logarithms.
Thanks in advance.

What is the derivation for the derivative of $a^t$
Been driving me nuts. Can someone prove to me that
$\frac{d(a^t)}{dt}=a^t\ln <!--\; \; -->(a)$
Thank you!

A logarithm integral
Calculate the integral
$\begin{array}{r}{\int <!--\; \int \; -->}_0^1\frac{\ln <!--\; \; -->(\sqrt{x}-<!--\; -\; -->\sqrt{1-<!--\; -\; -->x})}{\sqrt{x}}dx\end{array}$
and show the value is negative.

Why does this equation work?
let $P(x):=\underset{p\le <!--\; \le \; -->x}{\sum <!--\; \sum \; -->}Log[p]$, then we have
$P(2^{k+1})=\underset{i=0}{\overset{k}{\sum <!--\; \sum \; -->}}(P(2^{i+1})-<!--\; -\; -->P(2^i))<2\cdot <!--\; \cdot \; -->Log[2]\cdot <!--\; \cdot \; -->(1+2+4+...+2^k)\le <!--\; \le \; -->4\cdot <!--\; \cdot \; -->Log[2]\cdot <!--\; \cdot \; -->2^k$
Why does this equation work?
The first step is clear for me, since we have a telescope sum. However, how can I conclude to the first $<$ and the $\le <!--\; \le \; -->$ ?

How can $\frac{1}{2}\log <!--\; \; -->(x)=\log <!--\; \; -->(\sqrt{x})$?
How can $\frac{\log <!--\; \; -->(x)}{2}=\log <!--\; \; -->\left(\sqrt{x}\right)$?
How would I come to this conclusion?

How to simplify ${\ln }^2<!--\; \; -->\left(x\right)+2\ln <!--\; \; -->x-<!--\; -\; -->3$
I dont know how to simplify ${\ln }^2<!--\; \; -->\left(x\right)+2\ln <!--\; \; -->x-<!--\; -\; -->3$
I dont know how to get $(\ln <!--\; \; -->(x)+1)(\ln <!--\; \; -->(x)+3)$
But I am stuck and don't really know how to do that.
I tried something like this: $2\ln <!--\; \; -->x+2\ln <!--\; \; -->x-<!--\; -\; -->3,2\ln <!--\; \; -->\left(x(x-<!--\; -\; -->3)\right),2\ln <!--\; \; -->(x^2-<!--\; -\; -->3x)$

Tighter logarithmic inequality
There is a well-known lower bound for
$x\log <!--\; \; -->\frac{1+x}{x}\ge <!--\; \ge \; -->\frac{x}{1+x}$
for $x\ge <!--\; \ge \; -->0$. I know a tighter lower bound on the same domain
$x\log <!--\; \; -->\frac{1+x}{x}\ge <!--\; \ge \; -->\frac{2x}{1+2x}\ge <!--\; \ge \; -->\frac{x}{1+x}.$
It can be proved by Jensen's inequality. Do you know something tighter?