I absolutely adore your algebraic manipulative skills of how you got the two sides similar! However it's not the solution, and x cannot be real. I'll give you a tip: 1. Graph the two functions: 4^x and x in one xy-plane. 2. See if they intersect each other: in this case 4^x doesn't intersect x, since 4^x is growing exponentially and x linearly. So if two functions don't intersect each other, there's no real solution. You deserve a like for your awesome thinking.
There's no real solution because the function f(x) = ln(x)/x - ln (4) has no zero. You can see this immediately by drawing its graph. In th complex numbers you can take the square root of (-2) so you have solutions but you to modify your calculations by using i .
From the first step this is incorrect. You cannot 1/x on both sides of the function and keep to the real numbers. For example, for a value such as x=-2 the function (-2)^(-1/2) is undefined in the real numbers. The rest of the process is a series of errors resulting from this unfounded assumption.
@@Subhasistudu94 It may be sufficient to say then that this equation has no solutions on the real numbers. A bit of graphing or reasoning should suffice for this conclusion. The solutions are easy to find if you're familiar with the complex expansion of the logarithm but if you're not it may be quite a bit of reading.
@@Subhasistudu94 maybe we can use maclaurin series to expand 4^x and lets say we expand it till degree 2 then the value of x should be -0.2+-i and then the answer would be approximately equal to x but not exactly equal so we can further use higher degrees for which the answer might come out to be equal
I absolutely adore your algebraic manipulative skills of how you got the two sides similar!
However it's not the solution, and x cannot be real.
I'll give you a tip:
1. Graph the two functions: 4^x and x in one xy-plane.
2. See if they intersect each other: in this case 4^x doesn't intersect x, since 4^x is growing exponentially and x linearly. So if two functions don't intersect each other, there's no real solution.
You deserve a like for your awesome thinking.
Yes, i will definitely do so.
Thank you for correcting me. 🥺
There's no real solution because the function f(x) = ln(x)/x - ln (4) has no zero. You can see this immediately by drawing its graph. In th complex numbers you can take the square root of (-2) so you have solutions but you to modify your calculations by using i .
Thank you for correcting me. 🤐😶
From the first step this is incorrect. You cannot 1/x on both sides of the function and keep to the real numbers. For example, for a value such as x=-2 the function (-2)^(-1/2) is undefined in the real numbers. The rest of the process is a series of errors resulting from this unfounded assumption.
Can you please provide a possible solution ? (Pls)
Are you familiar with the lambert W function?
Not so familier
But i will surely look into it
Thank you for coreecting and watching...
🩷🤧😊
@@Subhasistudu94 It may be sufficient to say then that this equation has no solutions on the real numbers. A bit of graphing or reasoning should suffice for this conclusion. The solutions are easy to find if you're familiar with the complex expansion of the logarithm but if you're not it may be quite a bit of reading.
😪😊
Nice equation
Excellent 😊😊
Very nice🎉🎉❤
teria q aplicar a função w de Lambert
Sim, senhor.
obrigado por assistir
Wah sir ji🎉🎉🎉🎉🎉❤😊❤😊❤😊😊.
Bohot badiya
Aise videos late rahen😊
Keep Going.
Helpful videos🎉😊❤
wrong! If you draw the 2 curves y=x and y=4^x you will see that they don’t cross each other so no real roots
Thank you for the correction.
🩷🩷
Thanks for watching.
Johar Gomke
x = e ^ ( - W ( - ln 4 ))
No real roots
≈ 0.06396 - 1.09084 * i
Thank yoh for thr correction 🩷
i think this function 4^x-x will never be zero.
Thank you sir
For correcting me. 🥺
love it!
Thanku🩷
👌🏻👌🏻
🥺🤧
Good😆
আরো ভিডিও চাই ❤️
Nischoi
Want more videos 😊😶🌫️
Sure🤐
if x=-1/2 , 4^(-1/2) would be 1/2 and at RHS x would be -1/2 , hence it is not the solution
You are right.
But, according to all the rules used in this solution I am also not wrong.
So, what's the solution?
@@Subhasistudu94 maybe we can use maclaurin series to expand 4^x and lets say we expand it till degree 2 then the value of x should be -0.2+-i and then the answer would be approximately equal to x but not exactly equal so we can further use higher degrees for which the answer might come out to be equal
@crabbypatty7096 🥺🥺
1:37 ここで1/-(1/2)乗、すなわち-2乗しているから式が成り立たないと思います。
訂正してくれてありがとう
🥹🥺
4⁽⁻⅟²⁾ ≠ -1/2
logo 1/2 não é solução
Obrigado por corrigir.
Obrigado por assistir. 🩷
Isigaba sobudlelwane begazi
siyeza isigaba segazi sthandwa cc
Dear Subhasis Tudu: The Lines X and 4^X can not cross eachother, it means that the Answer MUST BE A COMPLEX NUMBER.
Thank you for correcting me... 😔🩷😮💨🥰
X= 0,06395
X=1,0908
Thank you🩷
Ø.
😱😳
Wrong, 4^-0.5 = 0.5 , which is not the solution. This equation has no solution.
Thank you for correcting.... 🤐
Very wrong, the solution is a complex one x=0.0476+1.1i approximately
Thank you for correcting me. 🥺