Oxford University Admission Test Tricks ! | ✍️🖋️📘💙

[@matheducare]

Very helpful math to learn. ❤

[@superacademy247]

I’m glad you found it helpful! 💯🙏🤩💕Thanks for sharing your support! 🔥🔥✅💕

[@davidseed2939]

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…)

[@ManojkantSamal]

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

[@motivaeducacao]

Excelent

[@christopherward2748]

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.

[@superacademy247]

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. 😁🙏💯

[@НеллиПшено]

Решаем методом устного счета. 3^х/9=3 3^x=27 x=3 Good luck!

[@2012tulio]

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

[@에스피-z2g]

3^x=9x 3=9x^(1/x) 3=27^(1/3) x=3