2^K = 10^X, 求解 K = X * (LN(10) / LN(2))

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tassadar
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2^K = 10^X, 求解 K = X * (LN(10) / LN(2))

Post by tassadar » 2019-04-23, 11:30

2^K = 10^X, 求解 K = X * (LN(10) / LN(2)) 求 K 為多少?

簡單答案: K = X * 3.321928

泛用公式 A^B = C^D,求解 B

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B = C * (LogN(D) / LogN(A))
舉例解法:

2^K == 10^50

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K == 50 * (LogN(10)/LogN(2))

LogN(10) == 2.3025850929940456840179914546843642076011014886288
LogN(2)  == 0.69314718055994530941723212145817656807550013436026

LogN(10)/LogN(2) == 3.3219280948873623478703194294893901758648313930246

We got K
K == 50 * 3.3219280948873623478703194294893901758648313930246
K == 166.09640474436811739351597147446950879324156965123

算出結果 Result:

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(2^166.09640474436811739351597147446950879324156965123) 
== 1.0000000000000000000000000000000000000000000000006E+50

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