問2008年的台灣入學試題1條(數學科試題1組)

By the question , we can set

2^-23=10^-k (a)
=>-23 log 2 = -k+log a
=>-6.9236899002715674899159945786633=-k+log a
Since log a should be a positive number,
so k must be >-6.9236899002715674899159945786633
=> k =7 ,
-6.9236899002715674899159945786633+7=log a
log a=0.07631009972843251008400542133666
a=1.1920928955078124999999999999982

so a =1//



2008-12-29 11:39:59 補充：
不過我憺心你明唔明我講咩.....
BECAUSE I MA POOR PERSENTATION...

唔該晒~我明白喇! ^V^

作答者答案雖然正確，但過程中卻有點疏忽

因為
2^-23=10^-k (a)並不是題目的意思

而開始應set

2^-23=(10^-k)( a+x/10) 而x是大或等於零而少於10的數字）