at 11:17 you have two expressions involving Lambert functions. but in the region we are considering, the Lambert function is dual valued W_0 andW_-1 it is from this fact that we get the second value for x. but you don’t describe the derivation of the number introduced at the end (0.12…)
I'm glad you found the W function useful! It's a great tool for solving this kind of problem. 💪🔥💯Thanks for sharing your approach! I bet you would be able to justify your solution to the examiner. 😁🙏💯
You can obtain the second solution by applying Lambert also : 3^x =9x ----->eq1 Let y = 3^x Take Ln both sides : x = ln y/ ln3 x = 0.91024 ln y ---->eq2 By substitution in eq 1 : y =9* 0.91024 ln y y = 8.192153 ln y ln y / y = 0.122068 y^-1 ln y^-1 = -0.122068 W(e^lny^-1 * ln y^-1 )= W(-0.122068) ln y^-1 = W(-0.122068) y^-1 = e^ W(-0.122068) y = 1/ e^ W(-0.122068) y = 1.150825 By substitution in eq2 : x = 0.91024 ln( 1.150825) x = 0.12787
Very helpful math to learn. ❤
I’m glad you found it helpful! 💯🙏🤩💕Thanks for sharing your support! 🔥🔥✅💕
at 11:17 you have two expressions involving Lambert functions.
but in the region we are considering, the Lambert function is dual valued W_0 andW_-1
it is from this fact that we get the second value for x.
but you don’t describe the derivation of the number introduced at the end (0.12…)
X=3, As per Lambert w
^=read as to the power
*=read as square root
As per question
3^(x-2)=x
3^x/3^2 =x
3^x/9=x
3^x=9x
Take log
Nice use of W function. But kinda did it in my head just going through the possibilities. I bet that the examiner would fail me for doing that.
I'm glad you found the W function useful! It's a great tool for solving this kind of problem. 💪🔥💯Thanks for sharing your approach! I bet you would be able to justify your solution to the examiner. 😁🙏💯
Excelent
By inspection in about 5 seconds…. X = 3.
Решаем методом устного счета.
3^х/9=3
3^x=27
x=3
Good luck!
You can obtain the second solution by applying Lambert also :
3^x =9x ----->eq1
Let y = 3^x
Take Ln both sides :
x = ln y/ ln3
x = 0.91024 ln y ---->eq2
By substitution in eq 1 :
y =9* 0.91024 ln y
y = 8.192153 ln y
ln y / y = 0.122068
y^-1 ln y^-1 = -0.122068
W(e^lny^-1 * ln y^-1 )= W(-0.122068)
ln y^-1 = W(-0.122068)
y^-1 = e^ W(-0.122068)
y = 1/ e^ W(-0.122068)
y = 1.150825
By substitution in eq2 :
x = 0.91024 ln( 1.150825)
x = 0.12787
3^x=9x
3=9x^(1/x)
3=27^(1/3)
x=3
3^x=9x , x*ln3=ln(9x) , /*9 , 9*x*ln3=9*ln(9x) , ln3/9=ln(9x)/(9x) , ln3/9=ln(9x)*e^(-ln(9x)) , /*(-1) , -ln3/9=-ln(9x)*e^(-ln(9x)) ,
((3/3)*(-ln3/9))=-ln(9x)*e^(-ln9x) , W(-3*ln3*e^(-3*ln3))=W(-ln(9x)*e^(-ln(9x))) , -3*ln3=-ln(9x) , -ln27=-ln(9x) , 27=9x , x1=3 , test1 , 3=3 , OK ,
W(-ln3/9)=-ln(9x) , x2=e^(-W(-ln3/9))/9 , x2=~ 0.127896 , test2 , 0.127896=0.127896 , OK ,