✔ 最佳答案
1.(log2 x)^log2 x=x之解為?
Set y = log2(x) = log(x)/log(2)......(1)=> log(x) = y*log(2) = log(2^y)=> x = 2^y = y^y=> y = 2 = log(x)/log(2)......by (1)=> log(x) = 2*log(2) = log(4)=> x = 4
2.某人向銀行貸款20萬元,約定從貸款後兩個月起,每兩個月均償還x元,分十年還清,假設每兩個月固定利率都是2.4%,每兩個月複利一次計算,則x之值為?(元以下不計,log4.15=0.6180)
n = 10*12/2 = 60x = sum*r*(1+r)^n / [(1+r)^n-1].....Excel: x=PMT解說= 200,000*0.024*1.024^60 / (1.024^60 - 1)= 200,000*0.024*4.15 / (4.15 - 1)......Note= 6324 $
Note: y = 1.024^60 => log(y) = 60*log(1.024)= 60*log(1024/1000)= 60*[log(1024) - log(1000)]= 60*[10*log(2) - 3*log(10)]= 60*(10*0.30103 - 3)= 60*0.0103= 0.6180=> y = 4.15
3.若織女星的亮度為x,則一顆亮度為y的星球,其星等p定義為:p=-2.5log(x/y)今已知天狼星為-1.4等星,北極星為2等星,求天狼星的亮度為北極星的幾倍?(log2.29=0.3598,log2.30=0.3617) -1.4 = -2.5*log(x/y) => 1.4/2.5 = log(x/y)=> x/y = 10^0.56
2 = -2.5*log(x/Y) => -2/2.5 = log(x/Y)=> x/Y = 10^(-0.8)
相除: Y/y = 10^1.36 = 10*10^0.36
Set z = 10^0.360.36 = log(z) = log[2.29 + (2.30-2.29)(0.36-0.3598)/(0.3617-0.3598)]= log(2.291053)=> z = 2.291053=> Y/y = 22.91 倍
2015-02-22 07:04:34 補充:
修改星等p定義為: p=2.5*log(x/y)
因為 y>x 才會出現負值
所以答案改為: y/Y = 22.91 倍
2015-02-22 19:15:04 補充:
第1題補充:
Set y = log2(x) => x = 2^y
x = [log2(x)]^log2(x)
=> 2^y = y^y
=> y = +-2
=> x = 2^y
= 2^(+-2)
= 4 or 1/4
第2題補充: Excel函數
x = pmt(0.024,60,200000)
= 6324 $
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