Алгебра 11 класс Логорифмы..

Ответ
V [25^{1/log(6) 5} + 49^{1/log(8) 7}] =
= V[(5^2)^{log(5) 6} +(7^2)^{log(7) 8}] =
= V[5^{2*log(5) 6} +7^{2*log(7) 8}] =
= V[5^{log(5) 6^2} +7^{log(7) 8^2}] =
= V[6^2 +8^2] =
= V(36 + 64) =
= V100 = 10

log(1/3) {5-2x} > -2
log(1/3) {5-2x} > log(1/3) {(1/3)^(-2)}
5-2x > (1/3)^(-2)
5-2x > 3^2
5-2x > 6
2x < -1
x < -1/2