※ 引述《aaaasd ()》之銘言：
: 已知a、b、c三數成等比
: 且log b、log c、log a三數成等差 求公差?
: a b c
: 思考很久 卻不得其解
: 懇求各位數學高手解答 非常謝謝
b^2 = ac
=> 2 = log a + log c
b b
d = log c - (1/log a) = log c -[1/(2-log c)] ..(1)
b b b b
=> 2log b = log a + 1
c c
d = log a - (1/log b) = 2log b - 1 - (1/log b) ..(2)
c c c c
由(1)(2)令x= log b 得： x-[1/(2-x)] = 2/x -1- x
c
=> 2x+1 = 2/x + [1/(2-x)] => x(2x+1)(2-x) = 4-2x+x = 4-x
=> 2x^3-3x^2-3x+4 = 0 = (x-1)(2x^2-x-4)
x = 1 or (1+√33)/4 or (1-√33)/4
所以公差d = 0 or -3/2
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※ 編輯: Intercome 來自: 140.92.63.232 (12/17 11:44)