Помогите log20 120, log9 20=a, lg2=b

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Алгебра

Декабристыч

21 февраля, 04:25

Помогите log20 120, log9 20=a, lg2=b

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Анюся · 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)