Recent content by RChristenk

##5x-6=5x-6## is defined as an identity because it is true for all values of ##x##. My question is I can further simplify and arrive at ##0=0##, in which case no values of ##x## will work because the variable ##x## itself doesn't exist. Isn't this a contradiction? Or did I violate some rule...

Why is the original fraction always positive? I'm confused because by definition ##\sqrt{x^2}=|x|##. So ##x## itself could be a negative number. Hence in ##\sqrt{\dfrac{1}{2x^3y^5}}##, ##x,y## could be negative and since it's to the third and fifth power respectively, stay negative. Thanks.

Nope. But if ##x,y## can be negative, would this be correct: ##\dfrac{1}{|x||y^2|\sqrt{2xy}}\cdot \dfrac{\sqrt{2xy}}{\sqrt{2xy}}=\dfrac{\sqrt{2xy}}{2x|x||y^2|y}##. Is there a way to simplify ##2x|x||y^2|y##?

##\sqrt{\dfrac{1}{2x^3y^5}}=\dfrac{1}{\sqrt{2\cdot x^2 \cdot x \cdot y^2 \cdot y^2 \cdot y}}=\dfrac{1}{|x|\cdot |y|\cdot |y| \cdot \sqrt{2xy}}=\dfrac{1}{|x|y^2\sqrt{2xy}}## ##\Rightarrow \dfrac{1}{|x|y^2\sqrt{2xy}} \cdot \dfrac{\sqrt{2xy}}{\sqrt{2xy}}=\dfrac{\sqrt{2xy}}{|x|y^2 \cdot...

I've multiplied everything out on paper and got ##x=2, \dfrac{15}{4}##, which is correct. However multiplying directly is tedious and from observing this problem I suspect there is a simplification or trick that I missed...

As you can see my instinct is to just plough ahead with the given equations instead of doing substitutions. Is there any specific characteristics to equations that give a hint to which substitution is most likely to succeed or is this just a case of doing loads of problems and memorizing what...

##\Rightarrow \begin{cases} (\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}})^2=(4\dfrac{1}{4})^2\\ (\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}})^2=(16\dfrac{1}{4})^2 \end{cases}## ##\Leftrightarrow \begin{cases} \dfrac{x}{y}+\dfrac{y}{x}+2=\dfrac{289}{16}\\ \dfrac{x^2}{y}+\dfrac{y^2}{x}...

##y+3=3\sqrt{y+7}## Square both sides: ##\Rightarrow y^2+6y+9=9y+63## ##\Rightarrow y^2-3y-54=0## ##\Rightarrow (y-9)(y+6)=0## ##y=9, -6## But if you plug in ##y=-6## into the original equation, you get ##-3=3## . So it doesn't work. Why?

I'm looking at this free book online: https://archive.org/details/elementaryalgebr00hall

Thanks for your replies. Why is ##\log_e(1+x)## invalid for ##x \geq 1##?

##\log_e\sqrt{x^2-1}=\dfrac{1}{2}[\log_e[(x+1)(x-1)]]=\dfrac{1}{2}[\log_e(x+1)+\log_e(x-1)]##. ##\Rightarrow \log_e(x-1)=\log_e[x(1-\dfrac{1}{x})]=\log_ex+\log_e(1-\dfrac{1}{x})## We know: ##\log_e(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots##...

Using ##\log_e(n+1)-\log_en=2[\dfrac{1}{2n+1}+\dfrac{1}{3(2n+1)^3}+\dfrac{1}{5(2n+1)^5}+\cdots]## and setting ##n=1##, the result is ##\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+ \cdots##. I thought this was different from ##1-\dfrac{1}{2}+\dfrac{1}{3}- \cdots##.

By definition: ##\log_e(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}- \cdots ## ##(1)## Replacing ##x## by ##−x##, we have: ##\log_e(1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}- \cdots## By subtraction, ##\log_e(\dfrac{1+x}{1-x})=2(x+\dfrac{x^3}{3}+\dfrac{x^5}{5}+ \cdots)## Put ##...

How did you go from ##\dfrac{n!}{(r-1)!(n-r+1)!}## to ##\dfrac{n(n-1)(n-2)\cdots(n-r+2)}{(r-1)!}##? I am assuming ##n!=n(n-1)(n-2)\cdots (n-r+2)(n-r+1)(n-r)\cdots 1##. Would this be correct? Thanks.

##r^{th}## term from beginning: ##^nC_{r-1}a^{n-r+1}(2x)^{r-1}## For the ##r^{th}## term from the end, we first know there are a total of ##n+1## terms in this binomial expansion. Subtracting the (##r^{th}## term from the end) from the total number of terms, ##n+1##, results in ##n+1-r## which...