Math - More about Equations (Logarithm) 2?
回答 (1)
2020-09-19 5:13 am
✔ 最佳答案
log_a(ab)．log_b(ab) + log_a(a-b)．log_b(a-b)= 0
= log_a(b)(log_b(ab))^2 + log_a(b)(log_b(a-b))^2
= log_a(b)[(log_b(ab))^2 + (log_b(a-b))^2]
iff. log_b(ab) = 0 and log_b(a-b) = 0 (∵ log_a(b) ≠ 0)
iff. ab = 1 and a-b = 1
iff. a = b+1 and b^2+b-1 = 0
iff. a = (√5 +1)/2, b = (√5 -1)/2 (∵ b > 0)
收錄日期: 2021-04-24 08:01:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200918201739AAjN1og