log 8 (x)+ log по основанию корень из 2 числа x = 14. помогите

Ответ
log(x) 8 + log(V2) x = 14
log(x) 2^3 + log(2^(1/2)) x = 14
3log(x) 2 + 1/(1/2)*log(2) x = 14
3/log(2) x + 2log(2) x = 14
3 + log^2(2) x = 14*log(2) x
log^2(2) x - 14*log(2) x + 3 = 0
log(2) x = t
t^2 - 14t + 3 = 0
Дальше легко.
Если в условии не log(x) 8, а log(8) x, то:
log(8) x + log(V2) x = 14
log(2^3) x + log(2^(1/2)) x = 14
1/3 * log(2) x + 2*log(2) x = 14
log(2) x^(1/3) + log(2) x^2 = 14
log(2) x^(1/3)*x^2 = 14
log(2) x^(1/3+2) = 14
2^14 = x^(7/3)
2^(14*3) = x^7
2^(14*3/7) = х
x = 2^6 = 64