急 ! Logarithmic Function 9p1

2012-04-19 6:16 pm
請詳細步驟教我計以下幾條.
(不要綱址回答)


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回答 (3)

2012-04-19 6:41 pm
✔ 最佳答案
46) log √0.1 - log 0.01

= (1/2) log 0.1 - log 0.01

= (1/2) (-1) - (-2)

= 3/2

47) log (1/8) - 3 log 5

= - log 8 - 3 log 5

= - 3 log 2 - 3 log 5

= -3 (log 2 + log 5)

= -3 log 10

= -3

90) log (2x + 6) - log x = 1

log (2 + 6/x) = log 10

2 + 6/x = 10

6/x = 8

x = 3/4

97) (log x)2 - 3 log x2 + 8 = 0

(log x)2 - 6 log x + 8 = 0

(log x - 4)(log x - 2) = 0

log x = 2 or 4

x = 100 or 10000

99) log (2x + 1)/log (x - 7) = 2

log (2x + 1) = 2 log (x - 7)

(2x + 1) = (x - 7)2

x2 - 14x + 49 = 2x + 1

x2 - 16x + 48 = 0

(x - 12)(x - 4) = 0

x = 12 or 4 (rejected since x - 7 > 0)

101) 32x+1 - 31(3x) + 36 = 0

3(32x) - 31(3x) + 36 = 0

[3(3x) - 4] [(3x) - 9] = 0

3x = 4/3 or 9

x = [log (4/3)]/log 3 or 2
參考: 原創答案
2012-04-20 1:45 am
46
log √0.1 - log 0.01
=(1/2)log0.1-log0.01
=-1/2+2
=3/2

47
log(1/8)-3log5
=log1-log8-log125
=-log8-log125
=-(log8+log125)
=-(log1000)
=-3

90.
log(2x+6)-logx=1
log(2x+6)/x=1
(2x+6)/x=10
2x+6=10x
6=8x
x=3/4

97
(logx)^2-3logx^2+8=0
(logx)^2-6logx+8=0
logx=4 或 logx=2
x=10000 或 x=100

99
[log(2x+1)]/[log(x-7)]=2
log(2x+1)=log(x-7)^2
2x+1=(x-7)^2
2x+1=x^2-14x+49
x^2-16x+48=0
x=12 或 x=4(捨去)

101
3^(2x+1)-31(3^x)+36=0
3[3^(2x)]-31(3^x)+36=0
3^x=9 或 3^x=4/3
x=3 或 xlog3=log4/3
x=3 或 x=0.262
2012-04-19 6:31 pm
log 1/8 - 3log5
= log 1/8 - log 5^3
= log 1/8 - log 125
= log (1/8 / 125)
= log (1/1000)
= log 10^(-3)
= -3log 10
= -3

其他唔識
參考: knowledge


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