關於log的問題

( 1 ) 2 ^ ( x + 3 ) - 5 ( 2^ x ) = 3
( 2 ^ x )( 2^3 ) - 5 ( 2 ^ x ) = 3
( 2 ^ x )( 8 - 5 ) = 3
2 ^ x = 1
x log 2 = log 1
x = 0
2) 2 log x + log 20 = 2
log x^2 + log 20 = log 100
log 20x^2 = log 100
20x^2 = 100
x^2 = 5
x = sqr5
3) ( 2 - log 2 ) / log 125
= ( log 100 - log 2 ) / log ( 5^3 )
= log 50 / 3 log 5
= ( log 5 + log 10 ) / 3 log 5
= ( log 5 + 1 ) / 3 log 5
= 1 / 3 + 1 / 3 log 5


2007-11-17 00:06:33 補充：
1 / 3 + 1 / 3 log 5也可以表示為( log 2 / 3 log 5+ 2/3 )的, 因為( log 2 / 3 log 5+ 2/3 )= ( log 2 + 2 log 5 ) / 3 log 5= ( 1 - log 5 + 2 log 5 ) / 3 log 5 [ 因log 2 + log 5 = 1 ]= ( 1 + log 5 ) / 3 log 5= 1 / 3 + 1 / 3 log 5

1. 2 ^(x+3)-5(2^x)=3
(2^x)(2^3) - 5 (2^x) = 3
8 (2^x) - 5 (2^x) = 3
3 (2^x) = 3
2^x = 1
x = 0

2. 2logx+log20=2
log x^2 = 2 - log20
log x^2 = log100 - log20
log x^2 = log(100/20)
x^2 = 5
x = √5


3. 2-log2
__________
log125

= log100-log2
__________
log125

= log(100/2)
__________
log125

= log50
__________
log125

= log(2*5^2)
__________
log5^3

= log2 + 2log5
__________
3log5

= log2 + 2/3
______
3log5

1. 0
2. 5^(1/2)
3. log50/log125

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