Here is a shorter solution which give the same answer: x² = 4^x x²=2^(2x) taking ln of both sides and using log priperties we get 2lnx=2xln2 lnx=xln2 lnx/x = ln2 Rewrite x as e^(lnx) lnx/[e^(lnx)] = ln2 Using exponent properties transport denominator into numerator we get (lnx)[e^(-lnx)] = ln2 (-lnx) [e^(-lnx) ] = -ln2 Taking W lambert of bothe sides we get -lnx=W(-ln2) lnx=-W(-ln2) finally rasing e to the power of both sides we get x = e^[-W(-ln2)] Which is the same as the answer in the video because ln4 / 2 can simplified into ln 2 as follows ln 4/2 ln(2²) /2 2ln2 / 2 ln2 This method is simpler, i dont know why he didnt do it this way in the video
Yeah, he was going on the right path, but he skipped a step. x positive and real is a contradiction; you can see this from the graph, and it shows up in the math because the Lambert W function only gives complex values for argument < -e^-1. He implicitly went back and took the negative root of the absolute value, that gives a positive argument to the W function, and the result he has pops out.
Here is a shorter solution which give the same answer:
x² = 4^x
x²=2^(2x)
taking ln of both sides and using log priperties we get
2lnx=2xln2
lnx=xln2
lnx/x = ln2
Rewrite x as e^(lnx)
lnx/[e^(lnx)] = ln2
Using exponent properties transport denominator into numerator we get
(lnx)[e^(-lnx)] = ln2
(-lnx) [e^(-lnx) ] = -ln2
Taking W lambert of bothe sides we get
-lnx=W(-ln2)
lnx=-W(-ln2)
finally rasing e to the power of both sides we get
x = e^[-W(-ln2)]
Which is the same as the answer in the video because ln4 / 2 can simplified into ln 2 as follows
ln 4/2
ln(2²) /2
2ln2 / 2
ln2
This method is simpler, i dont know why he didnt do it this way in the video
The -(ln4)/2 can be simplified to -ln2 if it is not to be evaluated.
So you finally found the solution using the function's graphs? Where you explain how to solve the Lambert's function numerically?
Numerical analysis using Newton's Method or any other, preferably with fast calculations rather than fast convergence if you are doing it by hand.
Raise bs to power 1/2
Divide bs by 2ˣ
x2⁻ˣ = 1 . . . use lambert
This is a communicative expression: 2^4 = 4^2
{4x+4x ➖ }=8x^2 2^3x^2 1^1^1x^2:1x^2 (x ➖ 2x+1).
You are wrong, the expression e^(-W(-ln(4)/2)) cannot have a negative value.
But your answer x ≈ -0.641 is correct.
How did this happen?
Correct solution:
x^2 = 4^x
ln(x^2) = ln(4^x)
2*ln|x| = x*ln(4) ===> two cases
1st case: x > 0
2*ln(x) = x*ln(4)
ln(x)*x^(-1) = ln(4)/2 # ln(4)/2 = ln(2^2)/2 = 2*ln(2)/2 = ln(2)
ln(x)*e^ln(x^(-1)) = ln(4)/2
ln(x)*e^(-ln(x)) = ln(4)/2
-ln(x)*e^(-ln(x)) = -ln(4)/2
W(-ln(x)*e^(-ln(x))) = W(-ln(4)/2)
-ln(x) = W(-ln(4)/2)
ln(x) = -W(-ln(4)/2)
x = e^(-W(-ln(4)/2)) ===> -ln(4)/2 < -1/e ===> no real solutions
2nd case: x < 0
2*ln(-x) = x*ln(4)
ln(-x)*x^(-1) = ln(4)/2
-ln(-x)*x^(-1) = -ln(4)/2
ln(-x)*(-x)^(-1) = -ln(4)/2
ln(-x)*e^ln((-x)^(-1)) = -ln(4)/2
ln(-x)*e^(-ln(-x)) = -ln(4)/2
-ln(-x)*e^(-ln(-x)) = ln(4)/2
W(-ln(-x)*e^(-ln(-x))) = W(ln(4)/2)
-ln(-x) = W(ln(4)/2)
ln(-x) = -W(ln(4)/2)
-x = e^(-W(ln(4)/2)
x = -e^(-W(ln(4)/2)) ===> ln(4)/2 > 0 ===> 1 real solution
x = -e^(-W₀(ln(4)/2)) = -0.641185744504985984486200482114823666562820957191101755139698797...
Yeah, he was going on the right path, but he skipped a step. x positive and real is a contradiction; you can see this from the graph, and it shows up in the math because the Lambert W function only gives complex values for argument < -e^-1. He implicitly went back and took the negative root of the absolute value, that gives a positive argument to the W function, and the result he has pops out.
Can you use better pen
Why not simplify ln(4)/2 to ln(2)?
Yes je me suis fait la meme réflection ca m'a énormément perturbé du début à la fin