F.4 log

1) log√8-log√4/5
= log√23 – log√(8/10)
= log√23 – log√(23/10)
= 0.5log23 – 0.5log(23/10)
= 1.5log2 – 1.5log2 + log10
= 1

2) (log9+3log2)/(log24-log 1/3)
= (log32+3log2)/(log(23x3)+log 3)
= (2log3+3log2)/(log23 +log3+log 3)
= (2log3+3log2)/(3log2 +2log3)
= (2log3+3log2)/(2log3+3log2)
= 1

2006-11-26 19:53:14 補充：
第一題更正1) log√8-log√4/5 = log√23 – log√(8/10)= log√23 – log√(23/10)= 0.5log23 – 0.5log(23/10)= 1.5log2 – 1.5log2 + 0.5log10= 0.5

1)log√8-log√4/5
=log[ √8/√(4/5)]
=log[(8/(4/5))^(1/2)]
=log[(10)^(1/2)]
=1/2
because log[(10)^(1/2)] is asking 10^x=10^(1/2)


2)(log9+3log2)/(log24-log1/3)
=log(9*3^2)/log(24/(1/3))
=log72/log72
=1

1) log√8-log√4/5 =?

=log(√8/√4/5)
=log(√2*√5)
=log(√10)
=log10^1/2
=1/2log10
=1/2(1)
=1/2


2) log9+3log2/log24-log1/3 =?

=log9+log8/log24-log1/3
=log(9*8)/log(24*3/1)
=log72/log72
=1


希望幫到你啦