請詳列演算步驟，多謝幫忙！
log2 (y^2 – 6y) = 3 + log2 (1– y)
代數(一)，數學達人請進。
2010-09-25 8:24 pm
回答 (2)
2010-09-25 8:34 pm
✔ 最佳答案
As follows:圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/17-13.jpg
2010-09-29 19:25:33 補充：
For logarithm function, y=logx (base number a>0)
The domain of x is positive number, so when y=2, log_2(1-2)=log_2 (-1), which is undefined.
Hence, y=2 is rejected.
2010-09-25 9:59 pm
log_2 (y^2 - 6y) = 3 + log_2 (1-y)
log_2 (y^2 - 6y) = log_2 2^3 + log_2 (1-y)
y^2 - 6y = 2^3 (1-y)
y^2 - 6y = 8(1-y)
...
log_2 (y^2 - 6y) = log_2 2^3 + log_2 (1-y)
y^2 - 6y = 2^3 (1-y)
y^2 - 6y = 8(1-y)
...
收錄日期: 2021-04-13 17:32:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100925000051KK00549