推 rey9810 : 哇 難過沒有詳解 感謝！ 01/01 13:27
※ 引述《rey9810 (RaY)》之銘言：
: http://i.imgur.com/0TSjO1p.jpg
: 這題不知道該怎麼令
: 請提示我一點 感謝 新年快樂！
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: Sent from JPTT on my LGE LG-H788.
log(3x) - [log(3x)] = 3{log(x) - [log(x)]}
log(3x) = log(3) + log(x)
log(x) = z + a, z為非負整數, 0 <= a < 1
設k, q為正整數
-k <= log(3) + log(x) < -k+1
(10^(-k))/3 <= x < 10 * (10^(-k))/3
有下列情況
(1)10^(-k)/3 <= x < 10^(-k)
log(3) + log(x) + k = 3[log(x) + k + 1]
=> log(x^2) = log(3) - 2k - 3
=> x = √[30 * 10^(-2k) * 10^(-4)]
= √30 * 10^(-k-2)
不合
(2)10^(-k) <= x < 10 * (10^(-k))/3
log(3) + log(x) + k = 3[log(x) + k]
=> log(x^2) = log(3) - 2k
=> x = √[3 * 10^(-2k)] = √3 * 10^(-k) (k = 1, 2, 3...
設k為非負整數
k <= log(3) + log(x) < k+1
=> (10^k)/3 <= x < (1/3) * 10^(k+1)
(3) (10^k)/3 <= x < 10^(k+1)
log(3) + log(x) - k = 3[log(x) - k]
=> log(x^2) = log(3) + 2k
=> x = √3 * 10^k (k = 0, 1, 2...)
(4) 10^(k+1) <= x < (1/3) * 10^(k+1)
log(3) + log(x) - k = 3[log(x) - k - 1]
=> log(x^2) = log(3) + 2k + 3
=> x = √30 * 10^(k+1)
不合
總結
x = √3 * 10^(k), k = 整數
x^2 = 3 * 10^(2k), k = 整數
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