Lambert w function always needs building an appropriate base to yield the good outcome. It requires experiences and practices to adjust the basic formula accordingly.
Take natural logarithm of equation: 2ln(x)= x ln(5-√24), and rearrange to -ln(x)exp(-ln(x)) = -ln(5-2√6)/2 =ln(√3+√2). Apply the Lambert W and solve for x: x=exp(-W(ln(√3+√2))) = 0.5390764227609346635539122391266984401562862123877397926588215125
The Wolfram approximation is not explained. And even if it were explained it is only just that - an approximation . This problem has almost nothing to do with aptitude.
Lambert w function always needs building an appropriate base to yield the good outcome. It requires experiences and practices to adjust the basic formula accordingly.
Take natural logarithm of equation:
2ln(x)= x ln(5-√24), and rearrange to
-ln(x)exp(-ln(x)) = -ln(5-2√6)/2 =ln(√3+√2).
Apply the Lambert W and solve for x:
x=exp(-W(ln(√3+√2))) = 0.5390764227609346635539122391266984401562862123877397926588215125
Simpler to expand e^x in a Taylor series without need of Lambert.
It is possible to semplify a bit the calculations observing that 5 - sqrt(24) = (sqrt(3) - sqrt(2))^2
X = e^-W (-1/2* Ln (5-sqrt (24)) is a more simplified solution.
Same method but different form of root.
x^2 =(5 - √24)^x = (5 - 2√6)^x = ((√3 -√2 )^2)^x = (√3 -√2)^(2x)
2lnx = 2x*ln(√3 -√2) => lnx/x = lnx/e^lnx = lnx * e^(-lnx) = ln(√3 -√2)
=> -lnx * e^(-lnx) = -ln(√3 -√2) => -lnx = W(-ln(√3 -√2))
=> lnx = -W(-ln(√3 -√2)) => x = e^lnx = e^(-W(-ln(√3 -√2)))
Cambridge University Admission Aptitude Problem: x² = (5 - √24)ˣ; x =?
x ≠ 0; (x²)¹⸍²ˣ = [(5 - √24)ˣ]¹⸍²ˣ, x¹⸍ˣ = [(5 - 2√6)]¹⸍² = √(5 - 2√6)
2(5 - 2√6) = 10 - 2(2)√6 = (√6)² - 2(2)√6 + 2² = (√6 - 2)²
5 - 2√6 = (√6 - 2)²/2 = [(√6 - 2)/√2]² = (√3 - √2)²
x¹⸍ˣ = √(5 - 2√6) = √[(√3 - √2)²] = √3 - √2 = 0.318; x¹⸍ˣ = 0.318
Trial-and-error math solution:
x = 1: x¹⸍ˣ = 1 > √3 - √2 = 0.318; 1 > x > 0; In-between
x = 0.5: 0.5¹⸍⁰·⁵ = 0.25 < 0.318; 1 > x > 0.5, Close to 0.5
x = 0.54: 0.54¹⸍⁰·⁵⁴ = 0.319 > 0.318; 0.54 > x > 0.5, Slightly < 0.54
x = 0.539: 0.539¹⸍⁰·⁵³⁹ = 0.318 = 0.318; Proved
The calculation was achieved on a smartphone with a standard calculator app
Answer check:
x = 0.539: x² = (5 - √24)ˣ; Confirmed as shown
Final answer:
x = 0.539
The Wolfram approximation is not explained. And even if it were explained it is only just that - an approximation . This problem has almost nothing to do with aptitude.
2*lnx=x*ln(5-V24) , lnx/x=ln(5-V24)/2 ,
W(-ln(5-V24)/2)=W(-lnx*e^(-lnx)) , 0.617897932=-lnx , x=e^(-0.617897932) , x=~ 0.539076 ,
test , x^2=0.290603 , (5-V24)^x=0.290603 , same , OK ,