F.4 log數 ! help

2010-09-11 10:06 pm
請詳細解答 !!
求過程 !

題1 )
12^2X-1=8^3X

題2)
log(6X-8)-log2X=log2

題3)
log(2X+6)-logX=1

題4)
logX^2+log2X=log16

題5)
logX^2=log2X+log3

6 )
log(2X^2-19)=log(5X-16)

7 )
2X+1=2.5105X
怎樣計下去 ??

感謝 ~!!!!

回答 (2)

2010-09-12 12:52 am
✔ 最佳答案
題1 )
12^2X-1=8^3X
log12^2x-log10=log8^3x
2xlog12-log10=3xlog8
2xlog12-3xlog8=log10
x(2log12-3log8)=log10
x=log10/(2log12-3log8)
x=-1.82

題2)
log(6X-8)-log2X=log2
log(6x-8/2x)=log2
log(3x-4/x)=log2
3x-4/x=2
3x-4=2x
x=4

題3)
log(2X+6)-logX=1
log(2x+6/x)=log10
2x+6/x=10
2x+6=10x
6=8x
x=4/3

題4)
logX^2+log2X=log16
log(x^2*2x)=log16
2x^3=16
x^3=8
x=2

題5)
logX^2=log2X+log3
logx^2=log(2x*3)
x^2=6x
x=6

6 )
log(2X^2-19)=log(5X-16)
2x^2-19=5x-16
2x^2-5x-3=0
(x-3)(2x+1)=0
x=3 or -1/2 (rejected)

7 )
2X+1=2.5105X
1=0.5105x
x=1.96


參考: me
2010-09-12 2:24 am
題1 )
12^(2X-1)=8^(3X)
(2x-1)log12 = 3xlog8
x(2log12-3log8) = log12
x=log12/(log9-log32)
x = -1.96


收錄日期: 2021-04-13 17:30:51
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