請問logx^2<log4x-3 則x=多少 log的底數是2?
回答 (1)
2019-01-17 10:58 pm
x² > 0 ∧ 4x - 3 > 0
x > 3 / 4 .
log₂ x² < log₂ (4x - 3)
x² < 4x - 3
x² - 4x + 3 < 0
(x - 1)(x - 3) < 0
1 < x < 3 .
1 < x < 3 #
x > 3 / 4 .
log₂ x² < log₂ (4x - 3)
x² < 4x - 3
x² - 4x + 3 < 0
(x - 1)(x - 3) < 0
1 < x < 3 .
1 < x < 3 #
2019-01-17 5:36 pm
log2_x^2<log2_(4x－3) 則x=?
Sol
log2_(4x－3)
4x－3>0
x>3/4
log2_x^2<log2_(4x－3)
x^2<4x－3
x^2－4x+3<0
(x－3)(x－1)<0
1<x<3
x>3/4
1<x<3
Sol
log2_(4x－3)
4x－3>0
x>3/4
log2_x^2<log2_(4x－3)
x^2<4x－3
x^2－4x+3<0
(x－3)(x－1)<0
1<x<3
x>3/4
1<x<3
收錄日期: 2021-04-30 22:47:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20190117082645AAn7z1q