Помогите log20 120 , log9 20=a, lg2=b

Question

Помогите log20 120 , log9 20=a, lg2=b

Answer 1

Log20 120 = log20 (20 * 6) =
=log20 20 + log20 6 =
=1 + (1 / 2) * log20 (6 ^ 2) =
=1 + (1 / 2) * log20 36 =
=1 + (1 / 2) * (log20 9 + log20 4)=
=1 + (1 / 2) * (1 / (log9 20) + 2 * log20 2)
=1 + 1 / (2a) + log20 2=
=1 + 1 / (2a) + 1/(log2 20)=
=1 + 1 / (2a) + 1/(log2 10 + log2 2)=
=1 + 1 / (2a) + 1/(1/(lg 2) + 1)=
=1 + 1 / (2a) + 1 / (1 / b + 1)

Comments

Спасибо