How To Solve x^2=-5Inx | Challenging Exponential Equation.

''How To Solve x^2=-5Inx", is a well detailed video on how to use the Lambert W function in solving tough exponential equation of this kind with natural logarithm.
In this video, Mr. Jakes took time to explain every step and how the mathematical laws/rules are applied to solving this challenging exponential equation to arrive at the two desired roots/solutions .
Watch from the beginning to the end without skipping any parts in order to get all the tricks/secrets therein.
Kindly share this nice and wonderful video tutorial with all those who need this knowledge in their field of endeavor using the video link below.
• How To Solve x^2=-5Inx...
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#howto #mathchallenge #tough #exponential #maths #toughproblems #lambert #wf #function

[@erfun5702]

Thanks for your teaching 👌

[@idontcare4715]

Perfect 👌🏼

[@onlineMathsTV]

Thanks for watching our videos and dropping this encouraging comment ma.
We all @OnlinemathsTV love you dearly ma 💖💖❤️❤️💕💕

[@sans1331]

x^2=-5lnx
raise e to both sides
e^x^2=(e^lnx)^-5=x^-5
raise both sides to 1/x^2 power
e=x^-5/x^2
raise both sides to -1/5 power
e^-1/5=x^1/x^2
im gonna try to use my proof for a similar problem with x^1/x instead of 1/x^2 to help get a picture on how to solve this
let x^(1/x^2)=y
x^(1/x^2)=y
y^x^2=x
lny^x^2=lnx
(x^2)lny=lnx
divide both sides by x^2
lny=(lnx)/x^2=(1/x^2)lnx=(x^-2)lnx
x^-2=e^ln(x^-2)=e^-2lnx
=(lnx)(e^-2lnx)=lny
*-2 both sides
-2lnx*e^-2lnx=-2lny
-2lnx=W(-2lny)
-2lnx=lnx^-2
raise e both sides
e^-2lnx=x^-2
x^-2=1/x^2=e^W(-2lny)
x^2=1/e^W(-2lny)=e^-W(-2lny)
x=sqrt(e^-W(-2lny))
if x^(1/x^2)=y,
x=sqrt(e^-W(-2lny)
overall conclusion:
if x^(x^n)=y
x=(e^W(nlny))^n
=> if x^(1/x^2)=e^-1/5
x=sqrt(e^-W(-2ln(e^-1/5)))
=sqrt(e^-W(2/5))

[@danielfranca1939]

Wow! This is a very nice solution. Well done Jakes....👍👍👍

[@onlineMathsTV]

Thanks a million @Danniel Franca for watching our content and at the same time commenting.
Much love ma...❤️❤️💕💕💕

[@ioannismichalopoulos6936]

In the original equation, you have lnx, thus x must be positive. As such, the negative solution should be rejected

[@rayrocher6887]

Thanks teacher, good equation, I like math genius

[@OnosuethaJamesJeremiah]

I like your mathematical level of explanation bro, thanks for this ability and thanks for making me understand the Lambert W function in a simple way. Mor grace from Above son.

[@onlineMathsTV]

Much appreciated sir and thanks for watch our contents

[@SidneiMV]

Great! Congrats sir.
x = ± √(e^-W(2/5)
another form is
*x = ± √[(5/2)W(2/5)]*
because e^W(x) = x/W(x)
what comes from W(x)e^W(x) = x
But I believe the negative value should be rejected because of ln(x) in the equation.

[@NitinKumar-hl8ud]

Thanks respected sir, love from India

[@nicogehren6566]

very nice question

[@kylekatarn1986]

Nice, but remember, since we have a logarithm, the argument must be positive, so x>0, we have to exclude the negative solution :)

[@onlineMathsTV]

Yap...thanks a million sir.

[@gracyledsy356]

What grade his teaching?

[@kylekatarn1986]

@@gracyledsy356 you mean, mine? High School from Italy

[@gracyledsy356]

@@kylekatarn1986 nice to meet u, I mean this video his teaching for what grade?

[@kylekatarn1986]

@@gracyledsy356 I think this is for challenges, like the ones you should find during the Olimpyades

[@chadiahamadou1478]

🙏🙏🙏

[@onlineMathsTV]

Thanks a million for watching our video tutorial and dropping this wonderful symbol of encouragement ma.
We all @onlinemathstv love you dearly ❤️❤️💖💖💕💕💕

[@chuksnonso2320]

Respect 🙏🙏🙏

[@onlineMathsTV]

Thanks for appreciating our little effort.
Also thanks for watching and dropping this encouraging comment.
Much love from everyone of us @onlinemathsTV 💕💕❤️❤️💖💖

[@hogec_enlightenmentarena6975]

great

[@onlineMathsTV]

Thanks for watching and leaving this comment sir.
Much love from everyone @onlinemathsTV 💕💕💕

[@keithturvey5818]

Thank you for introducing me to the Lambert W function. I am a retired former lecturer in physics to university students and have considerable interest in mathematical aspects of physics but, perhaps to my shame, confess that I have not met the Lambert function before- I don't think it appears in many of the well-known text books ( e.g. the widely recommended book by G Arfken) on mathematical methods for physics or similar books. You have given me a motivation to learn something about this function from online presentations.
By introducing the function f(x) = xsquared + 5 ln x and then finding the root of this function [i.e. setting f(x)=0] using the Newton-Raphson method of successive approximations I find the answer x=0.86193......
As you admit yourself, you did not finish with a numerical value for x but instead gave the result in terms involving the Lambert function. I find it puzzling that your final result for x in this presentation was +/- a particular value because if we agree that x=0.86193... is indeed one solution then how can x=- 0.86193... be another solution given that this would mean that the right-hand side of the original equation would then be complex but xsquared on the left hand side would still be real? The comment by @kylekatarn1986 essentially makes the same point.

[@onlineMathsTV]

So sorry the reply to this comment of yours is coming a bit late sir.
Your answer is correct from the Newton-Raphson method you applied.
As for me, I make use of the WolframAlpha calculator to get my numerical values from this point/stage of the math.
As for the minus sign, it will give positive answer also after the square root evaluation and so it will come out as 0.86193....because (-)×(-)=+.
Remember we took the square root of both sides of the equation @9:55
Thanks so much for watching and at the same time dropping this wonderful comment sir.
Much love and respect ❤️❤️💖💕💕🙏🙏🙏

[@billaynes8144]

Now the only thing to do is to run a computer program to approximate the numerical value of x, since the "W" function is only a definition...but of course we may have special purpose calculators to do the...

[@onlineMathsTV]

👍👍👌👌👌

[@hokie6384]

How do you get a numerical value?

[@keithturvey5818]

@hokie6384 One way to get a numerical value for x is to use a technique of successive approximations, e.g the Newton -Raphson method. I obtain x=0.86193.... although one should first find an approximate result before applying that method. In my case I initially found that x=0.5 and x=1.0 lie on opposite sides of the correct answer. For anyone adept at writing computer programs to numerically solve equations ( which I am not ) it would be an easy task to write a suitable successive approximations program.
You may find my comment elsewhere amongst the responses of interest.
I struggled a bit with Mr Jakes accent but I think he indicated that in a future online presentation he would show us how to get a numerical result- perhaps it will be more elegant than my way.

[@onlineMathsTV]

Use the WolframAlpha calculator sir.

[@opulence3222]

Not clear to me

[@ayhanguler.]

Yanlış 2/5 değil -2/5 olmalı

[@onlineMathsTV]

Sorry I did not get your point here sir, kindly point out the error again in a clearer form. Thanks sir.

[@corneliusagu2903]

Sorry, x can't be negative

[@onlineMathsTV]

ok

[@servetiskol8949]

False solve