Explore Fraction Concepts with Examples and Questions

Monty Hall variation
The question is a variation of the classic Monty Hall problem. Should one switch their first choice of door one, if there are four doors, with two cars and two goats, the second door containing a goat. How would one solve this using fractions?

$4{\log }_2<!--\; \; -->(n)=n$ How to Solve for $n$?
As the title suggests, my log skills are pretty lacking. Need to learn how to get from $4{\log }_2<!--\; \; -->(n)=n$ to $n=16$ ($\log$ base $2$). I've searched Google and it seems I am missing some core concept here. I really appreciate the help or guidance.

Find $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\left(\frac{x^2+1}{x^2-<!--\; -\; -->1}\right)}^{x^2}$
How to calculate the following limit?

Integration of $\int <!--\; \int \; -->\frac{1}{x^4+1}\mathrm{d}x$

Simple fraction calculation
I'm trying to write this expression as one fraction:
$\begin{array}{rl}\frac{5}{y+1}\cdot <!--\; \cdot \; -->\frac{1}{5}+\frac{y}{\frac{y+1}{3}}& =\frac{5y}{5y+5}+\frac{3y}{y+1}\\ & =\frac{5y^2+5+15y^2+15y}{5y^2+10y+5}\\ & =\frac{20y^2+15y+5}{5y^2+10y+5}\\ & =\frac{4y^2+3y+1}{2y}.\end{array}$
Am I doing this in the right way? Any hints about what to do next, please?

Pre-Algebra Fractional Exponent Question
Why does $t^{\frac{3}{2}}\cdot <!--\; \cdot \; -->t^{\frac{1}{2}}=t^2$?
What I tried to do was multiply the exponents together $\frac{3}{2}\cdot <!--\; \cdot \; -->\frac{1}{2}=\frac{3}{4}$ so my final answer was $t^{\frac{3}{4}}$ but according to my Wileyplus homework it is $t^2$.
Can someone please explain?

Proving question, pattern and inequalities
Hi my math tutor gave me this problem to do over the week: prove 1/1^2+1/2^2+1/3^2...1/1000^2 <2
I've managed to almost complete the question, but I think I'm missing something or made a mistake.
My attempt at the question: 1/1000^2 < 1/999*1000 1/2^2+1/3^2+...1/1000^2<1/1*2+1/2*3+...1/999*1000 1/1*2+1/2*3+...1/999*1000= (1-1/2)+(1/2-1/3)+(1/3-1/4)...+1/999-1/1000 (we can cancel out each fraction except for 1-1/1000)
Therefore, 1/1*2+1/2*3+...1/999*1000=1-1/1000 1/2^2+1/3^2+...1/1000^2<1-1000 1/1^2+1/2^2+1/3^2...1/1000^2<2-1000
As you can see, I'm stuck please help D;

How to put a fraction in simplest form, such as ${\displaystyle \frac{140}{255}}$?

rationalize the denominator
Is there any theorem saying that in below fraction we can't rationalize the denominator
$\frac{2}{\mathrm{\pi <!--\; \pi \; -->}}$
I couldn't find any way and I don't think there is but I was wondering if this actually proved? for example for $\frac{1}{\sqrt{2}}$ we say
$\frac{1}{\sqrt{2}}=\frac{1\times <!--\; \times \; -->\sqrt{2}}{\sqrt{2}\times <!--\; \times \; -->\sqrt{2}}=\frac{\sqrt{2}}{2}$
result have meaning if we look at fraction like Division

Sum of series with binary parity in the numerator
I'm now stuck with this question, and I don't even know where to start: Find sum of series
$\underset{1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{f(n)}{n(n+1)}$
, where f(n) - number of ones in binary representation of n.
I wish I could post some moves, that I've tried but I don't know what to do.
Thanks!

Finding x with other variables involved
How do you find $x$ when there are also other variables like $P,A,t$
$P=\frac{A}{1+xt}$
Here's what I did, I'm not sure if I'm correct
LCD: 1 + xt
$1+xt(P=\frac{A}{1+xt})$
$(1+xt)P=1+xt(\frac{A}{1+xt})$
$P+Ptx=A$
Is the above correct? I couldn't continue it because I don't know what to do next.

Solving equations including floor function.
I got a little trouble solving equations that involve floor function in an efficient way.
For example :
$\lfloor \frac{x+3}{2}\rfloor =\frac{4x+5}{3}$
In the one above, I get that you basically let
$\frac{4x+5}{3}=k$
and then inserting k in the left side, take $k=8l,8l+1$ and so on and test it.
If there's a better solution to the one above plese tell me.
My main problem is when it comes down to functions that have multiple floors such as :
$\lfloor \frac{x+1}{3}\rfloor +\lfloor \frac{2x+5}{6}\rfloor =\frac{3x-<!--\; -\; -->5}{2}$
Using the same method for each of them and then intersecting the solutions should give me the right answer but is there a faster way to solve equations like this ?

Can I take log on both sides of inequality such way?
$f(n)\le <!--\; \le \; -->c{n}^k$
and get
$\log <!--\; \; -->(f(n))\le <!--\; \le \; -->kc\log <!--\; \; -->n$
I know that by rules the result inequality should be like this
$\log <!--\; \; -->(f(n))\le <!--\; \le \; -->\log <!--\; \; -->(c{n}^k)$
but I read in book where inequality (1) $\to <!--\; \to \; -->$ in inequality (2) and want to know which way it was done
Edit: this is not general case, the questions relates to the algorithms and O-notation

If $\frac{a}{b}=\frac{c+d}{e+f}$ so $a$ is equal to $c+d$?
I have these question, a is always c+d and b is e+f ?.
Thanks.

How do I know that this is the LCD (Lowest Common Denominator)
When presented with the following expression :
$\frac{1}{(x+1)(x+2)}-<!--\; -\; -->\frac{2}{x+1}+\frac{3}{x+2}$
I have used an online calculator to show me the LCD, which is :
$(x+2)(x+1)$
Although I can work out the LCD for different algebraic expressions, I am unsure of the methodology behind solving this one. I realise it must be basic but it just has not clicked yet.
If anybody could explain how this is done I would be very grateful.

Find the value of $\frac{x_1^2}{1-<!--\; -\; -->x_1}+\frac{x_2^2}{1-<!--\; -\; -->x_2}+...+\frac{x_n^2}{1-<!--\; -\; -->x_n}.$.
Suppose, $x_1+x_2+...+x_n=1(x_i\in <!--\; \in \; -->\mathbb{R},x_i\ne 1)$ and $\frac{x_1}{1-<!--\; -\; -->x_1}+\frac{x_2}{1-<!--\; -\; -->x_2}+...+\frac{x_n}{1-<!--\; -\; -->x_n}=1.$ Find the value of $\frac{x_1^2}{1-<!--\; -\; -->x_1}+\frac{x_2^2}{1-<!--\; -\; -->x_2}+...+\frac{x_n^2}{1-<!--\; -\; -->x_n}.$

What is the right way of simplifying this trigonometric expression and why?
Problem:
$\frac{\frac{cosA}{sinA}-<!--\; -\; -->cosA}{\frac{cosA}{sinA}+cosA}$
sol1:
$=\frac{cosA(\frac{1}{sinA}-<!--\; -\; -->1)}{cosA(\frac{1}{sinA}+1)}$
$=\frac{(\frac{1}{sinA}-<!--\; -\; -->1)}{(\frac{1}{sinA}+1)}$
$=\frac{cosecA-<!--\; -\; -->1}{cosecA+1}$
sol2:
$=\frac{cosA-<!--\; -\; -->sinAcosA}{cosA+sinAcosA}$
$By÷cosA$
$=\frac{1-<!--\; -\; -->cosecA}{1+cosecA}$

Simplifying $\frac{3(a^{1/4}+4)}{2a-<!--\; -\; -->32a^{1/2}}$
I have a fraction $\frac{3a^{1/4}+12}{2a-<!--\; -\; -->32a^{1/2}}$ which I have factored out into $\frac{3(a^{\frac{1}{4}}+4)}{2a-<!--\; -\; -->32a^{\frac{1}{2}}}$ but checking out W|A I also get that there ought to be also $\frac{3}{2(a^{1/4}-<!--\; -\; -->4)a^{1/2}}$
I can't figure out what steps should I take between what I've got and that simpler form though.

Negative/Positive Index Numbers
I came across a question that I was unable to solve, it involves positive and negative index numbers in the same fraction and I'm not sure how to solve that, if anyone could help me please??
${\displaystyle \frac{\left(\frac{7a^5{b}^3}{5a^6{b}^2}\right)}{\left(\frac{7b^3{a}^2}{5b^5{a}^4}\right)}}$
Also some working out would be great so I can see how to work it out properly. Thanks!

Proving $\log <!--\; \; -->x^2=2\log <!--\; \; -->x$
How does $\log <!--\; \; -->x^2=2\log <!--\; \; -->x$
Can you do a proof please. I know that this is true but I don't know why.