Решить надо log8 128 + log4 16. 2) log3 9 –log9 - вопрос №69672368128416239927 от sultanguloverik2 28.11.2022 21:27

Question

Алгебра: Решить надо log8 128 + log4 16. 2) log3 9 –log9 27.

sultanguloverik2 · Answer

Log8 (128) + log 4(16) = 1/3 * log2(128) + 1/2* log2 (16) =
= log2 (128)^(1/3) + log2 (16)^(1/2) =
= log 2 (128^(1/3) * 4) =
= log 2 (2^(7/3) * 4) =
= log 2 ( 2^(7/3 + 2) ) =
= log2 ( 2^(7/3+6/3) ) =
= log 2 ( 2^(13/3) ) =
= 13/3

log3 (9) - log 9 (27) =
= log3 (9) - log3 (27)^(1/2) =
= log3 (9/√27) =
= log3 (3^(1/2)) =
= 1/2 = 0,5