Solving A Radical Exponential Equation

[@mrityunjaykumar4202]

he smartly skipped the 3rd power while doing it from 1st to 2nd then to 4th.. to make the video longer😆

[@SyberMath]

Ahahahahaha

[@sujitsivadanam]

I'm not quite sure what the point was of doing all that extra work just to show the cube of 1 + sqrt(3) is 10 + 6sqrt(3)

[@SyberMath]

So it’s an educated guess

[@NowInAus]

It was a technique and it led to a lan interesting pattern-based method using differences I’d squares. Yes it was long but it was educational.

[@sujitsivadanam]

@@NowInAus Thanks!!

[@jensraab2902]

Hey SyberMath! Have you considered making a video on this difference of two squares thing? I wasn't aware of this and I suppose I'm not the only one. 🙂

[@andylee63]

It's nice sometimes to digress a little and fiddle with the numbers even if it isn't the most direct path to the solution.

[@raminrasouli191]

Very good thank you. You are very through.

[@cameronspalding9792]

(1+sqrt(3))^3=10+6*sqrt(3), thus if u=(1+sqrt(3))^x, then we have u^3+u=10, by inspection u=2 is a root: thus (u-2) divides u^3+u-10, u^3+u-10 = (u-2)(u^2+2u+5)=0 The equation u^2+2u+5=0 has solution u= (-2+_sqrt(2^2-4*1*5))/2 = (-2+_sqrt(4-20))/2 = (-2 +_sqrt(-16))/2 = (-2 +_4i)/2 = -1 +_ 2i The solutions are given by x=log(2)/ln(1+sqrt(3)), x=log(-1+2i)/ln(1+sqrt(3)), x=log(-1-2i)/ln)1+sqrt(3)) where ln is defined as the single valued inverse of the exponential function that maps from R to R, log is the multi valued inverse of the complex exponential from C to C

[@haris4527]

Why we write ln(2)/ln(1+sqrt3) insted lg_(1+sqrt3)(2) (or log_(1+sqrt3)(2). )?

[@schukark]

Because ln is the "natural" log and it looks prettier that way(also, easier to type correctly), I guess that's the reasoning

[@bobbyheffley4955]

Log=base 10 or common logarithms; lg=base 2 or binary logarithms

[@giuseppemalaguti435]

10+6sqrt3=(1+sqrt3) ^3...x=log(1+sqrt3) 2...

[@barakathaider6333]

👍

[@s1ng23m4n]

easy solved by guessing 2^3 + 2 = 10 just need prove that (1+sqrt(3))^3 = 10 + 6*sqrt(3)

[@bsmith6276]

Yeah, I had similar thoughts. 10+6sqrt(3) feels like (1+sqrt(3))^3. Show that is true and the problem reduces quite quickly with the substitution y=(1+sqrt(3))^x. I'm a bit surprised at how long it took SyberMath to get there.

[@TDSONLINEMATHS]

Nice

[@popitripodi573]

Nice!!!❤🎉

[@SyberMath]

Thanks! 💕

[@scottleung9587]

Nice!

[@HoSza1]

not hard to solve if you already know the solution

[@SyberMath]

That’s true!

[@yhistory2688]

Can someone tell me to what audience this math is offered i mean to what math grade

[@yhistory2688]

@Hong Lyly thank you

[@keinKlarname]

In Germany it's University-level 😪

[@yhistory2688]

@@keinKlarname wow very comforting because i am a high schooler and i struggled with it thank you tick

[@yoav613]

Nice and easy,and it must be easy because otherwise you can not solve this!😃💯

[@wolfbirk8295]

You are wrong and you are right...

[@schukark]

First, observe, that (1+sqrt3)^3 = (10+6sqrt3). Then let t = (1+sqrt3)^x, equation becomes t^3+t=10. Let's introduce f(t) = t^3+t, it's increasing as it is a sum of 2 increasing functions(t and t^3). Then f(t) = const and f(t) is increasing gives us that f(t) = 10 has no more than one solution, by guessing it is 2. Then (1+sqrt3)^x=2 => x = ln(2)/ln(1+sqrt3). Problems like these are in our exam papers in 11th grade.