An Interesting Homemade Exponential Equation

[@MathOrient]

I appreciate your efforts to make up these equations :)

[@raymondarata6549]

Because the r.h.s. is sqrt(10)/2, x = 1/2 is a logical possible answer. In your video, you showed that sqrt(10)/2 = sqrt(5/2). Simply set 1 + 1/x = 5/2. Multiply both sides by x and then multiply both sides by 2 to get 2x^2 - 5x + 2 = 0. Solve it by factoring. You get (2x - 1)(x - 2) = 0 so x = 1/2 or 2. You check each answer and find that x = 2 is extraneous and x = 1/2 works! Knowing that the function on the l.h.s. is monotonic, you can be sure that x = 1/2 is the only solution.

[@maxborn7400]

math olympiad gigabrains right here

[@popitripodi573]

I loved it❤❤❤brilliant!!!

[@SyberMath]

Thank you! ❤❤❤

[@mehrdadbasiri9968]

Perfect...👍👍👍.

[@SyberMath]

Thanks 🤩

[@scottleung9587]

Nice!

[@vladimirkaplun5774]

An infinite source of the kitchen made equations. Replace 1/2 with 1/3 and another video is ready.

[@SyberMath]

Thanks for the idea! 😜😍😉

[@mcwulf25]

That was pretty much guess-and-check. Of course the essential bit is showing it's the only solution.

[@vbeckerlima]

What program do you use to make the resolution?

[@giordanociccoli]

its called desmos

[@carly09et]

nominalise the root to root(5/2). Then x as 1/2 is the trivial [(1/2) + 1/(1/2)]^(1/2)

[@hertselcorech9680]

Thank you! I always enjoy watching your shows. Can you please help me solve: X squared equals to 2 raised to the X. It should have 3 solutions but I don't know how to solve it.

[@abhinavanand9032]

There are only two real solutions x=2,4

[@michaelcampbell6922]

There are three real solutions, but only two integer solutions. x^2=2^x 2^2=2^2->x=2 2^4=(2^2)^2=4^2->x=4 These are the two integer solutions. Since there is an intersection on the left of the y-axis, the third solution must be negative. Let (-a)>0 be the absolute value of the third real solution: 2^(a)=(-a)^2=a^2 2^(a/2)=-a 2^(-a/2)=(-a)^(-1)=-1/a 2^[-(a+1)/2]=-1/(2a) -(a/2)*e^[-a*ln(2)/2]=1/2 [-a*ln(2)/2]*e^[-a*ln(2)/2]=ln(2)/2 -a*ln(2)/2=W[ln(2)/2], where W(x*e^x)=x a=-2W[ln(2)/2]/ln(2)

[@ParaNozek]

why (x+1/x)^x only defined when x>0?

[@SyberMath]

because that's when x+1/x > 0

[@ParaNozek]

@@SyberMath for example x=-2 => (-2-1/2)^-2 = 4/25 and it's positive. How can we exclude negative solutions? (I know there is none)

[@barakathaider6333]

👍

[@giuseppemalaguti435]

1/2

[@broytingaravsol]

x=1/2

[@MisterPenguin42]

"How do you come up with these equations?" I like to throw variables, integers, radicals, vincula, parantheses, and irrational numbers into a Nutribullet and then make a video. is what I'm expecting at the end of this video. Update: I was wrong.

[@SyberMath]

Ooopsies! Did I forget to talk about it?

[@MisterPenguin42]

@@SyberMath I don’t think so, but that just means more SyberMath content!!

[@MatthisDayer]

donald knuth is crying in the corner right now

[@mathswan1607]

x=1/2

[@rakenzarnsworld2]

x = 1/2