16^x² = x¹⁶ = (x²)⁸ x² = u 16ᵘ= u⁸ 2⁴ᵘ= (u²)⁴ 2ᵘ= u² uln2 = 2lnu (1/u)lnu = (1/2)ln2 (1/u)ln(1/u) = (1/2)ln(1/2) = (1/4)ln(1/4) => u = 2 or u = 4 u = 2 => x = ± √2 u = 4 => x = ± 2
@user-kp2rd5qv8gАй бұрын
Let t=x^2. Then, t ln16 = 8 ln t > 1/t ln t = 1/2 ln 2 > 1/t ln (1/t) = 1/2 ln (1/2) > ln (1/t) = W(1/2 ln(1/2)), where W is the Lambert W function. So t=e^[-W(1/2 ln(1/2))] > x = e^[-1/2 W(1/2 ln(1/2))] = √2, +/-2. So, x=√2, +/-2.