log對數問題

在 B 制下 ,N = log8 E
6.4 = log8 EE = 8^6.4
在 A 制下 , M = log4 E
M = log4 8^6.4
M = log4 (4*2)^6.4
M = log4 (4 * 4^0.5)^6.4
M = log4 (4^1.5)^6.4
M = log4 (4^9.6)
M = 9.6在 A 制下, 該爆炸的強度為 9.6.

2012-02-05 00:26:03 補充：
方法二 :

M / N

= log(4) E / log(8) E
= (log E / log4) / (logE / log8)
= log8 / log4
= log 2³ / log 2²
= 3 log2 / (2 log2)
= 3 / 2

M / 6.4 = 3 / 2

M = 9.6

2012-02-11 12:40:49 補充：
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Q) log對數問題


圖片參考：http://imgcld.yimg.com/8/n/HA07197115/o/701202040100413873434670.jpg
A)在 B 制下 ,N = log8 E
6.4 = log8 EE = 8^6.4
在 A 制下 , M = log4 E
M = log4 8^6.4
M = log4 (4*2)^6.4
M = log4 (4 * 4^0.5)^6.4
M = log4 (4^1.5)^6.4
M = log4 (4^9.6)
M = 9.6在 A 制下, 該爆炸的強度為 9.6. 方法二 :

M / N

= log(4) E / log(8) E
= (log E / log4) / (logE / log8)
= log8 / log4
= log 2³ / log 2²
= 3 log2 / (2 log2)
= 3 / 2

M / 6.4 = 3 / 2

M = 9.6 希望可以幫到你！

17
N=log E/log8
N=6.4
6.4=log E/log8
6.4log8=logE
E=8^6.4
所以
在A制上該爆炸的強度
M=log8^6.4/log4
M=9.6

∵ N= 6.4
∴ 6.4= log(8)E
8^(6.4)=E
∴In A, log(4)E=M
log(4)8^(6.4)=M
(6.4)(log8/log4)=M
M=6.4[(3xlog2)/(2xlog2)]
M=6.4x(3/2)
M=9.6
∴The magnitude in scale A is 9.6.