Помогите решить плиз
2log2 x^4+logx 2+9=0

2 log (2, x^4) + log (2, 2) / log(2, x) + 9 = 0 8 log (2, x) + 1 / log(2, x) + 9 = 0 8 log (2, x)^2 + 9 log(2, x) + 1 = 0 8 t^2 + 9t + 1 = 0 D = 49 = 7^2t1 = ( - 9 + 7)/16 = - 2/16 = - 1/8t2 = ( - 9 - 7)/16 = - 1 1) log (2, x) = - 1/8x = 2^(-1/8) ≈ 0,9172) log (2, x) = - 1x = 2^(-1)x = 0,5