[其他] log的連續計算
題目：
(1) log_10 20 + log_10 5 = 2
(2) log_8 12 + log_8 16 - log_8 3 = 2
小弟的想法：
(1) log_10 2*10 + log_10 (1/2)*10
= log_10 10 + log_10 2 + log_10 10 + log_10 (1/2)
= 1 + log_10 2 + 1 + log_10 (1/2)
= 2 + log_10 2 + log_10 (1/2)
= 2 + log_10 2*(1/2)
= 2 + log_10 1
= 2 + 1og_10 10^0
= 2 + 0*log_10 10
= 2 + 0*1 = 2
(2)log_2^(3) 4*3 + log_2^(3) 2^(4) - log_2^(3) 3
=log_2^(3) 2^(2) + log_2^(3) 3 + (4/3)log_2 2 - log+2^(3) 3
=(2/3)log_2 2 + (4/3)log_2 2 + log_2^(3) 3/3
=2*1 + log_2^(3) 1
=2 + (1/3)*log_2 1
=2 + (1/3)*0 = 2
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◆ From: 1.165.169.238
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02/03 18:49, , 1F
02/03 18:49, 1F
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02/03 18:56, , 2F
02/03 18:56, 2F
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02/03 19:41, , 3F
02/03 19:41, 3F
已修正！不好意思，小弟眼睛又脫窗了！
※ 編輯: pigheadthree 來自: 1.165.169.238 (02/03 19:55)
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