(√1995)·x^(log₁₉₉₅ x) = x²

[@narhumo7497]

Oh wow the mod was awesome, very well explained sir. This is a great problem to test algebraic skill. I really like how it wraps up to be 1995^2, it gets so simple from something so disturbingly complicated at first glance

[@nicolascamargo8339]

With square difference it is also possible to calculate 1995² quickly

[@dirklutz2818]

@@nicolascamargo8339 Indeed! 1995² = (2000-5)² = 2000² - 20000 + 25, so the last three digits are 025.

[@PrimeNewtons]

Smart

[@marcolima89]

this guy has genuinely rekindle my interest in math thank you for doing this and please keep doing it.

[@toastdog214]

I tried doing it myself and I got to 1995², and I was stumped on how to get the last 3 digits. I ended up just doing regular multiplication, but wow the solution you have given here is genius! mod 1000 would've never crossed my mind

[@nicolascamargo8339]

With square difference it is also possible

[@majora4]

What I love most about this problem is how generalizable it is. You can replace 1995 with any number y > 1 and the product of the positive roots will always be y^2. Moreover, you can always replace the exponent on the right-hand side with any number z > sqrt(2)* and the product of the positive roots will always be y^z. *If you set z = sqrt(2) there will be only one root... but it has multiplicity 2 and it value is exactly sqrt(y^z) so you can make an argument that the product should still be considered y^z

[@VictorZheng-sc5sc]

so beautiful😍😍

[@allozovsky]

We can add some nice Unicode typesetting to the title: *(√1995)·x^(log₁₉₉₅ x) = x²*

[@allozovsky]

I don't know if the "vinculum" in *√1̅9̅9̅5̅* would be visible on all devices, though 🤔

[@PrimeNewtons]

Thank you

[@ritwikgupta3655]

LOL, nothing scares you, my man! Well done!

[@Moj94]

I tried doing it myself and I decided I'd rather stay an infant.

[@souverain1er]

Awesome Prof.!

[@davidgagen9856]

Brilliant explanation.

[@mudspud]

Fascinating

[@souverain1er]

I think the last 3 digits of the product of the positive roots is 1995^2 (mod 1000), which equals (1995 mod 1000)^2 using modular exponent property. You skipped a step.

[@Happycurv]

I'm impressed with how far I got. The question confused me I guess though

[@SiladityaSen1993]

Isnt doing the (2000-5)^2 be easier?

[@YAYAMEDIA-su8up]

I love you so much and am from mom Ethiopia

[@spartan246]

This god is a goddam wizzard

[@Arkapravo]

weird question, cool solution!

[@PauloDacosta-s1s]

Do you have a video explain mod theory and exercises?

[@the_real_nayak]

As an Indian, we do this in 11th STD

[@TaiserBinJafor]

The answer is 025.

[@wiafesamuel-xi8cl]

Please help me these questions. x to the power x equal to 3 to the power 81 another one is x to the power x equal to 2 to the power 64 and lastly, x to the power x equal to 2 to the power 1024.

[@jamesharmon4994]

Since you have two roots, both positive, wouldn't you have three possible products? (Root 1)^2, (Root2)^2, (Root1)(Root2)??

[@nicolascamargo8339]

Dato curioso y sí puede ser

[@JEEAspirant1738]

sir I didn't understand the part where you put mod 1000. Please explain it to me in the comments.

[@JCSol]

crazy

[@Ron_DeForest]

How is it you can just say the base is 1995? Isn’t the base automatic 10? If you’re changing the base, don’t you have to do shit to the argument? Am I writing this correctly?

[@vojtatheguitargeek]

yeah he could have already written it in the first step...he basically took log base 1995 of both sides of the equation, but firstly thought of using regular log and then he decided to change the base afterwards and didn't come back to rewrite it everywhere. as far as i know you don't have to mess with the argument if you choose whatever base you like (as long as you apply the same log to both sides).

[@knupug]

The base of the log function can be anything. It simply asks what power of the base will equal the argument. The ln function is log base e. Log to the base 2 of 8 is 3. Log to the base 10 of 100 is 2. So, to solve this problem, he took log to the base 1995 of both sides.

[@sphakamisozondi]

If u introduce log, u can use any base of your choice. If u want to do it the long way around. U can use base 10, then do change of Base, and u will arrive to the same approach but with extra steps. He just skipped two steps in order to use base 1995. But if u are comfortable with the base 10 approach then apply change of Base procedure, u can do so.

[@aguyontheinternet8436]

if y=x, then log(y)=log(x), regardless of the base you choose. You can go back and change the base as many times you want, the equation will still hold true

[@Adam-pj2qh]

wow thats clever