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(1/2)^x = x^(1/2)taking log, (x)log(1/2) = (1/2)log(x) => log(1/2)/(1/2) = log(x)/(x) => x=1/2
Just compare the base and the power
Very nice! ❤
(1/2)^x=x^(1/2)(1/4)^x=x1/4=x^(1/x)x=1/2
Very good
Thanks ❤
Japanese, Math Olympiad: (1/2)ˣ = x¹⸍²; x =?x > 0, Let: a = 1/2(1/2)ˣ = aˣ = x¹⸍² = xᵃ, xᵃ = aˣ, (xᵃ)¹⸍⁽ᵃˣ⁾ = (aˣ)¹⸍⁽ᵃˣ⁾, x¹⸍ˣ = a¹⸍ᵃ; x = a = 1/2Answer check:x = 1/2: (1/2)ˣ = (1/2)¹⸍² = x¹⸍²; ConfirmedFinal answer:x = 1/2
(1/2)^x = x^(1/2)
taking log, (x)log(1/2) = (1/2)log(x)
=> log(1/2)/(1/2) = log(x)/(x)
=> x=1/2
Just compare the base and the power
Very nice! ❤
(1/2)^x=x^(1/2)
(1/4)^x=x
1/4=x^(1/x)
x=1/2
Very nice! ❤
Very good
Thanks ❤
Japanese, Math Olympiad: (1/2)ˣ = x¹⸍²; x =?
x > 0, Let: a = 1/2
(1/2)ˣ = aˣ = x¹⸍² = xᵃ, xᵃ = aˣ, (xᵃ)¹⸍⁽ᵃˣ⁾ = (aˣ)¹⸍⁽ᵃˣ⁾, x¹⸍ˣ = a¹⸍ᵃ; x = a = 1/2
Answer check:
x = 1/2: (1/2)ˣ = (1/2)¹⸍² = x¹⸍²; Confirmed
Final answer:
x = 1/2
Very nice! ❤