Explore Fraction Concepts with Examples and Questions

Find the Solution of the Exponential Equation?
How do I solve $5^x=4^x+1$?
I understand how to solve for $x$ when there is one exponent, but I don't know how to solve when there is an exponent on both sides of the equation..

Simplifying Rational Expressions
Simplify the following rational expression: $5/(x+3)-<!--\; -\; -->7x/(x-<!--\; -\; -->1)$
I came across this question in my homework and because it is a fraction, I decided that I needed to establish a common denominator of:
$x-<!--\; -\; -->3$
But how do I convert x - 1 into x-3 ?

Ordering an even and odd fraction that are close
We know that $1/4<5/11<1/2$
I did it this way from small to large:
$\frac{1\cdot <!--\; \cdot \; -->3}{4\cdot <!--\; \cdot \; -->3}=\frac{3}{12}$
$\frac{5}{11}$
$\frac{1\cdot <!--\; \cdot \; -->6}{2\cdot <!--\; \cdot \; -->6}=\frac{6}{12}$
It is hard for me to decide between 5/11 and 6/12 without a calculator. Is there an easier way to compare them?

Does there exist $a,b,c,d$ such that $\frac{a+b+c+d}{4}$ is an integer?
Let $a,b,c,d$ are different positive integers, such that a,b,c, nor d is a multiple of a different variable.
Question:
Does there exist $a,b,c,d$ for
$\frac{a+b+c+d}{4}=I$
such that I is an integer?

How to differentiate for critical points with variable in denominator
sorry for posting a particular problem, this is maybe more of an algebra problem than a calculus problem, but it does involve differentiating so I thought I would state the problem as one. I am currently running through the openOnlineMIT multivariable calculus course. I have ran into a problem which I can't seem to come to the solution. Specifically I am trying to find $\frac{\mathrm{d}}{\mathrm{d}t}\frac{4}{1+t^2+(1.5-<!--\; -\; -->t^2{)}^2}=0$ When I set this to zero I get $\frac{1}{t(1-<!--\; -\; -->t^2)}=0$. What I can't seem to understand how to solve the algebra. The answers given is $4t(t^2-1)=0$. which is a bit different but is solvable.

Simplify the expression: $\frac{3^{2n}-<!--\; -\; -->1}{3^{n+1}-<!--\; -\; -->3}$
Factorizing indices is quite a cool and intriguing area of indices. However, I stumbled upon one that I couldn't do any further:
$\frac{3^{2n}-<!--\; -\; -->1}{3^{n+1}-<!--\; -\; -->3}$
I got up to this stage:
$\frac{3^n\cdot <!--\; \cdot \; -->3^n-<!--\; -\; -->3^0}{3^n\cdot <!--\; \cdot \; -->3^1-<!--\; -\; -->3^1}$
I got to this stage through using the laws of indices and breaking them down. I feel that I'm close, but not there yet. Please help. Thank you!

How to prove that $\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\ge <!--\; \ge \; -->\sqrt{x}+\sqrt{y}$
I am trying to prove that
$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\ge <!--\; \ge \; -->\sqrt{x}+\sqrt{y}$
I tried some manipulations, like multiplications in $\sqrt{x}$ or $\sqrt{y}$ or using $x=\sqrt{x}\sqrt{x}$, but I'm still stuck with that.
What am I missing?

Solve the Logarithmic Equations for x, please.
This one is an exponential equation that I can't figure out..
$7^{x-<!--\; -\; -->2}=5^{3-<!--\; -\; -->x}$
These two are logarithmic equations that I'm also having trouble with..
$\ln <!--\; \; -->\sqrt[3]{x-<!--\; -\; -->6}=-<!--\; -\; -->2$
${\displaystyle \frac{1025}{7+e^{4x}}=5}$
These ones really stumped me. Any explanations would be greatly appreciated! Thank you!

How did my book go from: $\frac{4}{5}=\frac{x}{30}$ to $\frac{4}{1}=\frac{x}{6}$
I understand that I could have cross multiplied it in the first place but what I don't understand is why my book changes the denominators without changing the numerators when simplifying the problem.

Why is this true? $p_x$ $x$ $(1+\frac{(1-<!--\; -\; -->a)}{a})$ = $p_x{w}_x+p_y{w}_y$$\Rightarrow <!--\; \Rightarrow \; -->$$x$ = $a$$\frac{p_x{w}_x+p_y{w}_y}{p_x}$
I don't see why the following equation is true - although wolfram-alpha gives me the same result, I can't figure out the steps that were made. Sure, we can simply divide the equation by $p_x$, but what happens when we also divide $(1+\frac{(1-<!--\; -\; -->a)}{a})$?

Derivative with a fraction freaks me out
I have this:
$L(\mathrm{\mu <!--\; \mu \; -->},{\mathrm{\sigma <!--\; \sigma \; -->}}^2)=-<!--\; -\; -->\frac{1}{2{\mathrm{\sigma <!--\; \sigma \; -->}}^2}\underset{n=1}{\overset{N}{\sum <!--\; \sum \; -->}}(x_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}{)}^2$
and I need to end up here:
$\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{\mathrm{\sigma <!--\; \sigma \; -->}}^2}L(\mathrm{\mu <!--\; \mu \; -->},{\mathrm{\sigma <!--\; \sigma \; -->}}^2)=\frac{1}{2{\mathrm{\sigma <!--\; \sigma \; -->}}^4}\underset{n=1}{\overset{N}{\sum <!--\; \sum \; -->}}(x_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}{)}^2$
But when I try to work things out, I do:
$L(\mathrm{\mu <!--\; \mu \; -->},{\mathrm{\sigma <!--\; \sigma \; -->}}^2)=-<!--\; -\; -->2^{-<!--\; -\; -->1}({\mathrm{\sigma <!--\; \sigma \; -->}}^2{)}^{-<!--\; -\; -->1}\underset{n=1}{\overset{N}{\sum <!--\; \sum \; -->}}(x_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}{)}^2=-<!--\; -\; -->2^{-<!--\; -\; -->1}\underset{n=1}{\overset{N}{\sum <!--\; \sum \; -->}}(x_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}{)}^2({\mathrm{\sigma <!--\; \sigma \; -->}}^2{)}^{-<!--\; -\; -->1}$
thus
$\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{\mathrm{\sigma <!--\; \sigma \; -->}}^2}L(\mathrm{\mu <!--\; \mu \; -->},{\mathrm{\sigma <!--\; \sigma \; -->}}^2)=\frac{1}{2}\underset{n=1}{\overset{N}{\sum <!--\; \sum \; -->}}(x_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}{)}^2{\mathrm{\sigma <!--\; \sigma \; -->}}^2$
What am I missing? :/

Spivak Calculus 3rd. Edition Chapter 1 Problem 12 (iii) and (iv) Proofs Critique
Here are my "proofs" for Spivak's Calculus Chapter 1 Problem 12. I am new to this level of rigour and I am attempting to intimate myself with more advanced topics of mathematics to prepare for next year. I apologize in advance, as these so-called "proofs" are not likely to be nearly as rigorous as they should be. Any assistance on how to write the proofs better or any critiques on faulty logic would be greatly appreciated.
12) iii) Prove that $\frac{|x|}{|y|}=\left|\frac{x}{y}\right|$
My proof is:
$\frac{|x|}{|y|}=|x|\cdot <!--\; \cdot \; -->|y^{-<!--\; -\; -->1}|$, therefore by (i), $|x|\cdot <!--\; \cdot \; -->|y^{-<!--\; -\; -->1}|=|x\cdot <!--\; \cdot \; -->(y^{-<!--\; -\; -->1})|=\left|\frac{x}{y}\right|$
12) iv) Prove that $|x-<!--\; -\; -->y|\le <!--\; \le \; -->|x|+|y|$
My proof is:
$|x|+|y|\ge <!--\; \ge \; -->|x+y|$, replacing $y$ with $(-<!--\; -\; -->y)$ yields
$|x|+|-<!--\; -\; -->y|\ge <!--\; \ge \; -->|x+(-<!--\; -\; -->y)|=|x|+|y|\ge <!--\; \ge \; -->|x-<!--\; -\; -->y|$, which is what we wished to prove.

Limit $\underset{x\to <!--\; \to \; -->0}{lim}\frac{1}{\ln <!--\; \; -->(x+1)}-<!--\; -\; -->\frac{1}{x}$
The problem is I don't know if I can calculate it normally like with a change of variables or not. Keep in mind that I'm not allowed to use L'Hôpital's rule nor the $\mathcal{O}$-notation.

Tips and tricks for speedy mental division with harder fractions
I'm looking for a strategy to solve these kind of questions rapidly. Do you guys have any suggestions?
$\frac{\frac{15}{25}}{X}=\frac{14}{35}.$
A. $\frac{3}{5}$
B. $\frac{63}{30}$
C. $\frac{14}{15}$
D. $\frac{9}{6}$
The only thing I can think of, is to look at
$\frac{14}{35}.$
and notice that the denominator has increased compared to the numerator, hence I can rule out A and C. Leaving only 2 options.

Relative width of one image wrt another
Suppose I have an image:
120: Original Width
70: Original Height

I have another image:
20: Original Width
20: Original Height
Both these images have to be relative to each other (or should I say these both are in aspect ratio with each other?)
So what would be the height of second image if the height of first image decreases to 100?
Now I do know the way to get aspect ratio with changing height for one single image, which would be (if I am not wrong): Old Width/Old Height * New Width = New Height but how do I apply something like this to two images?

Exponential function negative: ${\left(\frac{81}{4}\right)}^{1/4}{\left(\frac{1}{4}\right)}^{-<!--\; -\; -->3/4}$
Multiply on both sides equals ${\displaystyle \frac{{81}^{1/4}}{4^{1/4}}}\cdot <!--\; \cdot \; -->{\displaystyle \frac{1^{-<!--\; -\; -->3/4}}{4^{-<!--\; -\; -->3/4}}}$
This should be ${\displaystyle \frac{3}{4^{1/4}}}\cdot <!--\; \cdot \; -->{\displaystyle \frac{1^{-<!--\; -\; -->3/4}}{4^{-<!--\; -\; -->3/4}}}$ which I get to -3/${\sqrt{4}}^{-<!--\; -\; -->1/2}$
Is that correct? I should get the answer 6.

Fractions from least to greatest
What is the fastest way to find the least common denominator of all the fractions without losing too much time?
7/9 , 1/4, 14/15, 2/3, 1/2
Thanks

World Problem Math Algebra Fraction
The denominator of a fraction in simplest form is greater than the numerator by $3$. If $7$ is added to the numerator, and $5$ added to the denominator, then the fraction itself is increased by ${\displaystyle \frac{1}{2}}$. Find the original fraction.
I got the equation ${\displaystyle \frac{x}{x+3}}={\displaystyle \frac{2x+14}{x+8}}$
When I solve this, I get a weird answer. My equation might be wrong. Can someone help me?

How to calculate $n$th term in terms of constants?
The expression is
$t_n=\frac{(x\times <!--\; \times \; -->t_{n-<!--\; -\; -->1}{)}^2}{((x-<!--\; -\; -->t_{n-<!--\; -\; -->1}\times <!--\; \times \; -->y{)}^2+4\times <!--\; \times \; -->x\times <!--\; \times \; -->t_{n-<!--\; -\; -->1})\times <!--\; \times \; -->t_{n-<!--\; -\; -->2}}$
where $x$ and $y$ are constants.
$t_0$,$t_1$, $t_2$, $t_3$ and $t_4$ are known.
How to calculate $t_n$?

Which rules are used to make function like one in Laplace Transformations table?
I have function like this:
$\frac{s^2+3s+3}{(2s^2+7s+7)}$
It needs to be brought to the level of Laplace Transformations from table, like these two:
$\frac{a}{(s-<!--\; -\; -->b{)}^2+a^2}$
$\frac{s-<!--\; -\; -->b}{(s-<!--\; -\; -->b{)}^2+a^2}$
I have the solution for this problem, but I can't figure out the logic behind it. Steps included are:
$\frac{1}{2}-<!--\; -\; -->\frac{0.5s+0.5}{2s^2+7s+7}=\frac{1}{2}-<!--\; -\; -->\frac{0.5s+0.5}{(s+1.75{)}^2+{0.6614}^2}=\frac{1}{2}-<!--\; -\; -->0.5\ast <!--\; \ast \; -->\frac{s+1.75}{(s+1.75{)}^2+{0.6614}^2}+(0.5\ast <!--\; \ast \; -->1.75-<!--\; -\; -->0.5)\ast <!--\; \ast \; -->\frac{1}{0.6614}\ast <!--\; \ast \; -->\frac{0.6614}{(s+1.75{)}^2+{0.6614}^2}$