Improve Your Algebra Skills with Practice Problems

Summing reciprocal logs of different bases
I recently took a math test that had the following problem:
$\frac{1}{{\log }_2<!--\; \; -->50!}+\frac{1}{{\log }_3<!--\; \; -->50!}+\frac{1}{{\log }_4<!--\; \; -->50!}+\cdots <!--\; \cdots \; -->+\frac{1}{{\log }_{50}<!--\; \; -->50!}$
The sum is equal to 1. I understand that the logs can be broken down into (first fraction shown)
$\frac{1}{{\log }_2<!--\; \; -->1+{\log }_2<!--\; \; -->2+{\log }_2<!--\; \; -->3+\cdots <!--\; \cdots \; -->+{\log }_2<!--\; \; -->50}$
How do the fractions with such irrational values become $1$? Is there a formula or does one simply need to combine fractions and use the basic properties of logs?

Hard question simplify as much as possible following
$\frac{x^3-<!--\; -\; -->49x}{x^4-<!--\; -\; -->13x^3+50x^2-<!--\; -\; -->56x}$
I get answer $\frac{x+3}{x-<!--\; -\; -->1}$ but my friend say it's wrong I dont understand please help me please

How do you solve ${\displaystyle -2+x\ge -9}$ ?

Let ${\displaystyle f\left(x\right)=3-(x+4)+2x}$ . How do you find all values of x for which f(x) is at least 6?

Let we have $\mathrm{\xi <!--\; \xi \; -->}$ which has normal distribution $N(a,{\mathrm{\sigma <!--\; \sigma \; -->}}^2)$
And also let's define $\mathrm{\eta <!--\; \eta \; -->}=e^{\mathrm{\xi <!--\; \xi \; -->}}$. I want to find $E_{\mathrm{\eta <!--\; \eta \; -->}}$ and $D_{\mathrm{\eta <!--\; \eta \; -->}}$ for this random value.
Here's what I've got:
I can find $E_{\mathrm{\xi <!--\; \xi \; -->}}$ and $D_{\mathrm{\xi <!--\; \xi \; -->}}$ because I know that for $\mathrm{\xi <!--\; \xi \; -->}$:
$p(x)=\frac{1}{\sqrt{2\mathrm{\pi <!--\; \pi \; -->}{\mathrm{\sigma <!--\; \sigma \; -->}}^2}}{e}^{\frac{(x-<!--\; -\; -->a{)}^2}{2{\mathrm{\sigma <!--\; \sigma \; -->}}^2}}$
and then we can evaluate that
$E_{\mathrm{\xi <!--\; \xi \; -->}}={\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}xp(x)dx$
Same things for disperson and I've got that
$D_{\mathrm{\xi <!--\; \xi \; -->}}=E_{{\mathrm{\xi <!--\; \xi \; -->}}^2}-<!--\; -\; -->E_{\mathrm{\xi <!--\; \xi \; -->}}^2=(a^2+{\mathrm{\sigma <!--\; \sigma \; -->}}^2)-<!--\; -\; -->a^2={\mathrm{\sigma <!--\; \sigma \; -->}}^2$
But how can I use it to evaluate $E_{\mathrm{\eta <!--\; \eta \; -->}}$ and $D_{\mathrm{\eta <!--\; \eta \; -->}}$?

Intuitively, why does ${\displaystyle \frac{a}{c}}={\displaystyle \frac{1}{\frac{c}{a}}}$?
For intuition, I reference objects. Imagine making a dessert with: $a$ as apples and $c$ as chestnuts.
Question. How and why is
${\displaystyle \frac{a}{c}}(3)={\displaystyle \frac{1}{{\displaystyle \frac{c}{a}}}}(4)?$
I ask only about intuition; please omit formal arguments and proofs (eg: Intuition is not generated by the explanation that rationalising (4)'s denominator produces (3)).

Let $X$ and $Y$ be two iid random variables on $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{F},\mathbb{P})$. I want to show that $X-<!--\; -\; -->Y$ has a symmetric distribution. I know that ${\mathbb{P}}_X(A)={\mathbb{P}}_Y(A)$ for every $A\in <!--\; \in \; -->\mathcal{B}(\mathbb{R})$. How can I show that ${\mathbb{P}}_{X-<!--\; -\; -->Y}(A)={\mathbb{P}}_{Y-<!--\; -\; -->X}(A)$ for every $A\in <!--\; \in \; -->\mathcal{B}(\mathbb{R})$? I know that it should be enough to analyze the family of intervals $(\mathrm{\infty <!--\; \infty \; -->},x]$, but I don't see how I can relate $\mathbb{P}(Y\le <!--\; \le \; -->x)=\mathbb{P}(X\le <!--\; \le \; -->x)$ to $\mathbb{P}(X-<!--\; -\; -->Y\le <!--\; \le \; -->x)=P(Y-<!--\; -\; -->X\le <!--\; \le \; -->x)$.

In AoPS' (Art of Problem Solving) proof of Muirhead's inequality, how does the below equality work out?
The below equation appears to show two expressions (1 and 2), each under the symmetric sum notation, being multiplied together to produce expression 3, which is also under the symmetric sum notation...
but normally,$[A_1+A_2(\dots <!--\; \dots \; -->)+A_n]\cdot <!--\; \cdot \; -->[B_1+B_2+(\dots <!--\; \dots \; -->)+B_n]$ is not $[A_1{B}_1+A_2{B}_2+(\dots <!--\; \dots \; -->)+A_n\cdot <!--\; \cdot \; -->B_n]$ ,
that is, when we multiply the sum of $A_i(i=[1,n])$ and $B_i(i=[1,n])$, the result is not the sum of $A_i\cdot <!--\; \cdot \; -->B_i(i=[1,n])$

I am reading a study in which the mean age of a population is 41 years. The mode and median are however both 35 years. I do not have access to the raw data.
How I can interpret the difference between the mean and the mode/median?

rational numbers as upper limit of a summation?
a quick question: Is it a legit way to use a fraction as the upper limit of a summation? Given is a frequency $f$ and a sample rate $f_s$. I want to use a sum like this:
$\underset{k=1}{\overset{\frac{f_s}{2f}}{\sum <!--\; \sum \; -->}}\sin <!--\; \; -->(2\cdot <!--\; \cdot \; -->\mathrm{\pi <!--\; \pi \; -->}\cdot <!--\; \cdot \; -->k)$
Is this a legit way to write it?

Josephine started a business selling cosmetics. She spent $4500 to obtain her merchandise, and it costs her S200 per week for general expenses. She earns $550 per week in sales. What is the minimum number of weeks it will take to make a profit?

Is it a common or accepted practice to specify measurement units directly in the formula?
Something like
$x_{mi}=\frac{y_{km}\star <!--\; \star \; -->123-<!--\; -\; -->z_{mm}}{456}$
where X is given in miles, Y in kilometers, and Z in millimeters.

Is 95 a ' number?

How do you find the ' factorization of 20?

Steps to simplify specific fraction
What are the steps to simplify the first fraction to the second?
$\frac{\frac{3-<!--\; -\; -->(3+h)}{3(3+h)}}{h}⟹<!--\; ⟹\; -->\frac{-<!--\; -\; -->1}{3(3+h)}$
I assume the first step would be to re-write it like so:
$\frac{3-<!--\; -\; -->(3+h)}{3(3+h)h}$
But after that I don't know where to go...

Measurements of some variable X were made at an interval of 1 minute from 10 A.M. to 10:20 A.M.. The data, thus, obtained is as follows:
X:60,62,65,64,63,61,66,65,70,68,63,62,64,69,65,64,66,67,66,64,50.
The value of $X$ which is exceeded 50% of the time in the duration of measurement, is
(i) 69
(ii) 68
(iii) 67
(iv) 66

Answer:
It is a multiple choice question and it is general knowledge question.
The answer is 68.
But I am unable to understand the question. What exactly says the question?

How do you solve and graph ${\displaystyle -1\le 5+b<10}$ ?

Preamble: Suppose that $m^{\ast <!--\; \ast \; -->}$ is the Radon outer measure on ${\mathbb{R}}^n$ and $S\subseteq <!--\; \subseteq \; -->{\mathbb{R}}^n$. Define $r_i^{\ast <!--\; \ast \; -->}(S)=(m^{\ast <!--\; \ast \; -->}|B_i(0))(S)=m^{\ast <!--\; \ast \; -->}(S\cap <!--\; \cap \; -->B_i(0)),S\subseteq <!--\; \subseteq \; -->{\mathbb{R}}^n$ the restriction of $m^{\ast <!--\; \ast \; -->}$ to the open ball of radius i centered at the origin. Assume it to be known that 1.) $r_i^{\ast <!--\; \ast \; -->}$ is a Radon outer measure, and $r_i^{\ast <!--\; \ast \; -->}({\mathbb{R}}^n)\le <!--\; \le \; -->m^{\ast <!--\; \ast \; -->}(\stackrel{¯<!--\; ¯\; -->}{B_i(0)})<\mathrm{\infty <!--\; \infty \; -->}$, 2.) for every $\mathrm{ϵ<!--\; ϵ\; -->}>0$ there exists closed set C and open set D such that $C\subseteq <!--\; \subseteq \; -->S\subseteq <!--\; \subseteq \; -->D$, $m^{\ast <!--\; \ast \; -->}(S\setminus <!--\; \setminus \; -->C)<\mathrm{ϵ<!--\; ϵ\; -->},m^{\ast <!--\; \ast \; -->}(D\setminus <!--\; \setminus \; -->S)<\mathrm{ϵ<!--\; ϵ\; -->}$, 3.) if the set S is $m^{\ast <!--\; \ast \; -->}$-measurable (by Caratheodory's criterion), then it is also $r_i^{\ast <!--\; \ast \; -->}$-measurable.
Then, there exists open sets $S\subseteq <!--\; \subseteq \; -->D_i$ such that $r_i^{\ast <!--\; \ast \; -->}(D_i\setminus <!--\; \setminus \; -->S)<\frac{\mathrm{ϵ<!--\; ϵ\; -->}}{2^{i+1}}$, $i=1,2,\dots <!--\; \dots \; -->$. Define $D=\underset{i=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\bigcup <!--\; \bigcup \; -->}}(D_i\cap <!--\; \cap \; -->B_i(0))$.
Main question: What I am struggling to understand is that why $S\subset <!--\; \subset \; -->D$? We certainly know that S belongs to every $D_i$. But what "guarantee" there is that S belongs to the intersection of $D_i$ with the open ball $B_i(0)$? My reading material does not provide any further information about this, and proceeds onwards with the surrounding proof about a property of Radon outer measures.

Inequality on non-negative reals
For non-negative $x,y,z$ satisfy $\frac{1}{2x^2+1}+\frac{1}{2y^2+1}+\frac{1}{2z^2+1}=1$ then show that Idea how to handle the constraint? I'm unaware .

What is $\frac{x^{10}+x^8+x^2+1}{x^{10}+x^6+x^4+1}$ given $x^2+x-<!--\; -\; -->1=0$?