數學 有關log的問題

(1)
1 + log4(x + 1) = log2(x - 7)
1 + [log10(x + 1) / log104] = log10(x - 7) / log102
(log102 / log102) + [log10(x + 1) / 2log102]= log10(x - 7) / log102
log102 + [log10(x + 1) / 2] = log10(x - 7)
2log102 + log10(x + 1) = 2log10(x - 7)
log10[4(x + 1)] = 2log10(x - 7)
4(x + 1) = (x - 7)²
4x + 4 = x² - 14x + 49
x² - 18x + 45 = 0
(x - 3)(x - 15) = 0
x = 3(不合、捨棄，因 log2(x - 7) =log2(-4)) 或 x = 15


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(2)
log102x + log­104 = 3
log10(2x*4) = log101000
8x = 1000
x = 125


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(3)
log6(x² + x) < 1 及 log6(x²+ x) 有定義
log6(x² + x) < log66 及 x² + x >0
x² + x < 6 及 (x + 1)x> 0
x² + x - 6 < 0 及 (x + 1)x> 0
(x + 3)(x - 2) < 0 及 (x + 1)x> 0
-3 < x < 2 及 (x < -1 或 x > 0)
-3 < x < -1 及 0 < x< 2


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(4)
log0.5(x² + 3x) > -2 及 log0.5(x²+ 3x) 有定義
log0.5(x² + 3x) > -2log0.5(0.5) 及 x² + 3x > 0
log0.5(x² + 3x) > log0.5(0.5)⁻² 及 (x + 3)x > 0
log0.5(x² + 3x) > log0.5(4) 及 (x + 3)x > 0
log(x² + 3x) / log0.5 > log4 / log0.5 及 (x + 3)x> 0

由於 log0.5 < 0 :
log(x² + 3x) < log4 及 (x + 3)x> 0
x² + 3x < 4 及 (x + 3)x> 0
x² + 3x - 4 < 0 及 (x + 3)x> 0
(x + 4)(x - 1) < 0 及 (x + 3)x > 0
(-4 < x < 1) 及 (x < -3 或 x > 0)
-4 < x < -3 及 0 < x < 1

第一題其實沒有換成以10為底的必要性。

直接用log_4(4x + 4) = log_4 (x - 7)^2
因為底數4 不等於 1，因此4x + 4 = (x - 7)^2