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))
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.
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.
Thanks for appreciating our little effort. Also thanks for watching and dropping this encouraging comment. Much love from everyone of us @onlinemathsTV 💕💕❤️❤️💖💖
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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.
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 ❤️❤️💖💕💕🙏🙏🙏
@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.
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...
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Thanks teacher, good equation, I like math genius
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very nice question
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))
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.
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.
Much appreciated sir and thanks for watch our contents
In the original equation, you have lnx, thus x must be positive. As such, the negative solution should be rejected
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Also thanks for watching and dropping this encouraging comment.
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Nice, but remember, since we have a logarithm, the argument must be positive, so x>0, we have to exclude the negative solution :)
Yap...thanks a million sir.
What grade his teaching?
@@gracyledsy356 you mean, mine? High School from Italy
@@kylekatarn1986 nice to meet u, I mean this video his teaching for what grade?
@@gracyledsy356 I think this is for challenges, like the ones you should find during the Olimpyades
great
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Thanks a million for watching our video tutorial and dropping this wonderful symbol of encouragement ma.
We all @onlinemathstv love you dearly ❤️❤️💖💖💕💕💕
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.
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 ❤️❤️💖💕💕🙏🙏🙏
How do you get a numerical value?
@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.
Use the WolframAlpha calculator sir.
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...
👍👍👌👌👌
Not clear to me
Yanlış 2/5 değil -2/5 olmalı
Sorry I did not get your point here sir, kindly point out the error again in a clearer form. Thanks sir.
Sorry, x can't be negative
ok
False solve