thank you!
更新1:
Thank you for you answering:) But for step 2 & 3, I guess 4^(2x)≠16^x. Then the solution is wrong, isn't it?
7(4^(2x)-1) = 12^(x+3)?
回答 (3)
2011-12-31 9:34 pm
✔ 最佳答案
If your question haven't typing error ,7(4^(2x) - 1) = 12^(x+3)
4^(2x) - 1 = (12^3)/7 * 12^x
16^x - 1 = (1728 / 7) * 12^x
(16/12)^x - 1 / 12^x = 1728 / 7
(4/3)^x - (1/12)^x = 1728 / 7
As (1/12)^x is very small , so
(4/3)^x ≈ 1728 / 7
x log (4/3) ≈ log (1728 / 7)
x ≈ log(1728 / 7) / log(4/3)
x ≈ 19.149
2012-01-05 23:00:50 補充：
4^(2x) = 16^x is no problem.
a^(mn) = (a^m)^n
So 4^(2x) = (4^2)^x = 16^x
2012-01-04 1:39 am
4^(2x)
= (4^x)(4^x)
= (4^x)^2
= (4^2)^x
= 16^x
= (4^x)(4^x)
= (4^x)^2
= (4^2)^x
= 16^x
2011-12-31 8:41 am
I think you better use graphical method.
收錄日期: 2021-04-21 22:22:59
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111230000051KK00612