log 的問題

100 [ (1+a)²⁴- 1 ] = 20000a
(a+1)²⁴- 1 = 200a
(a+1)²⁴- 1 + 200 = 200a + 200
(a+1)²⁴- 200(a+1) = - 199
(a+1) [ (a+1)²³ - 200 ] = - 199情況一 :
當 a+1 < 0 及 (a+1)²³ - 200 > 0
得 a < - 1 及 a > ²³√200 - 1 ≈ 0.25906
茅盾!情況二 :
當 a+1 > 0 及 (a+1)²³ - 200 < 0
得 a > - 1 及 a < ²³√200 - 1
==>
- 1 < a < ²³√200 - 1 ≈ 0.25906再看 (a+1) [ (a+1)²³ - 200 ] = - 199當 - 1 < a < 0 時 , - 199 = (a+1) [ (a+1)²³ - 200 ] < (0+1) [ (0+1)²³ - 200 ] = - 200 茅盾!當 a = 0 時 , (a+1) [ (a+1)²³ - 200 ] = - 199 成立
得 a = 0 為一解。 當 0 < a < ²³√200 - 1 時 ,
則 (a+1)²³ - 200 = - 199 / (a+1) < - 199 / ²³√200 ≈ - 158.05502
即
(a+1)²³ - 200 < - 158.05502
a < 0.17639利用分半方法 :設 f(a) = (a+1)²⁴- 200(a+1) + 199 = 0試驗易知f(0.15) < 0
f(0.16) > 0
得 0.15 < a < 0.16f( (0.15 + 0.16)/2 ) = f(0.155) < 0
得 0.155 < a < 0.16f( (0.155 + 0.16)/2 ) = f(0.1575) > 0
得 0.155 < a < 0.1575f( (0.155 + 0.1575)/2 ) = f(0.15625) > 0
得 0.155 < a < 0.15625f( (0.155 + 0.15625)/2 ) = f(0.15563) > 0
得 0.155 < a < 0.15563f( (0.155 + 0.15563)/2 ) = f(0.15532) < 0
得 0.15532 < a < 0.15563f( (0.15532 + 0.15563)/2 ) = f(0.15548) < 0
得 0.15548 < a < 0.15563f( (0.15548 + 0.15563)/2 ) = f(0.15556) > 0
得 0.15548 < a < 0.15556f( (0.15548 + 0.15556)/2 ) = f(0.15552) > 0
得 0.15548 < a < 0.15552故 a = 0.1555 (四位小數)
圖像見 :
http://www.wolframalpha.com/input/?i=%28a%2B1%29^24-+200%28a%2B1%29+%2B199%3D+0+

https://lh3.googleusercontent.com/-SpMCbrj6Onc/ThuhE_rIHJI/AAAAAAAAGbI/wCBGNaLma2s/0r1555x.JPG


圖片參考：https://lh3.googleusercontent.com/-SpMCbrj6Onc/ThuhE_rIHJI/AAAAAAAAGbI/wCBGNaLma2s/0r1555x.JPG