关于log的问题
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( 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 补充： 1 / 3 + 1 / 3 log 5也可以表示为( log 2 / 3 log 5+ 2/3 )的
因为( 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
参考: My Maths Knowledge
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 应该对吧 ~ 希望帮到你
因为( 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
参考: My Maths Knowledge
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 应该对吧 ~ 希望帮到你
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