[數學]求a.b?
Re: [數學]求a.b?
由 Searchtruth 於 星期二 四月 01, 2003 9:46 pm
索洛西 寫到:已知方程式 l logx l=ax+b有3個實根，其比為1:2:3，求a=? b=?
已知方程式 l logx l=ax+b有3個實根，其比為1:2:3，求a=? b=?
設此根為 t ,則 x 分別為 t , 2t , 3t t>0
當 ㏒t > ㏒1 = 0
原e.q.=>
㏒t =at+b
㏒2t=2at+b
㏒3t=3at+b
=>無解
當 -㏒2 < ㏒t < ㏒1 =0
原e.q.=>
-㏒t =at+b
㏒2t=2at+b
㏒3t=3at+b
=>t= √(3)/2
當 -㏒3 < ㏒t < -㏒2
原e.q.=>
-㏒t =at+b
-㏒2t=2at+b
㏒3t=3at+b
=>無解
當 ㏒t < -㏒3
原e.q.=>
-㏒t =at+b
-㏒2t=2at+b
-㏒3t=3at+b
=>無解
=>t 唯一解為 √(3)/2
=>-㏒√(3)/2=√(3)/2*a+b
㏒√(3)=√(3)*a+b
=>a=2√(3)/3*㏒3/2
b=㏒4√(3)/9
答案怪怪ㄉ說~"~可能算錯ㄌ...
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Searchtruth - 訪客
由 Searchtruth 於 星期二 四月 01, 2003 10:24 pm
無解...因為㏒xㄉ真數x必須為正
即三根必須同號~
請問一下喔~你為什麼會想到(-96):6:37這一組數字ㄚ?
很沒有規則喔~@@"
即三根必須同號~
請問一下喔~你為什麼會想到(-96):6:37這一組數字ㄚ?
很沒有規則喔~@@"
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Searchtruth - 訪客
由 Searchtruth 於 星期三 四月 02, 2003 6:03 pm
嗯?!真數可以為負ㄇ?!@@"我倒是沒有想過...不過~剛剛想想~~
應該不能為負ㄉㄅ?!
對ㄌ...原本題目ㄉ解答誰知道阿?~~謝謝拉~~~
應該不能為負ㄉㄅ?!
對ㄌ...原本題目ㄉ解答誰知道阿?~~謝謝拉~~~
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Searchtruth - 訪客
由 Raceleader 於 星期五 四月 04, 2003 7:43 pm
This is only my guess
a,b,k≠0
After testing, k≠1/3, 1/2, 1
If 0＜k＜1/3
logk=-(ak+b) ---(1)
log2k=-(2ak+b) ---(2)
log3k=-(3ak+b) ---(3)
(1)-(2):
log(1/2)=ak
(2)+(3):
log(2/3)=ak
Which is impossible
If 1/3＜k＜1/2
logk=-(ak+b) ---(1)
log2k=-(2ak+b) ---(2)
log3k=3ak+b ---(3)
(1)-(2):
log(1/2)=ak
(2)+(3):
log(6k2)=ak
6k2=1/2
k=1/√12 (Rejected)
If 1/2＜k＜1
logk=-(ak+b) ---(1)
log2k=2ak+b ---(2)
log3k=3ak+b ---(3)
(1)+(2):
log(2k2)=ak
(3)-(2):
log(3/2)=ak
2k2=3/2
k=√3/2
If 1＜k
logk=ak+b ---(1)
log2k=2ak+b ---(2)
log3k=3ak+b ---(3)
(2)-(1):
log2=ak
(3)-(2):
log(3/2)=ak
which is impossible
k=√3/2
After checking a, b value, it does not fix value
a,b,k≠0
After testing, k≠1/3, 1/2, 1
If 0＜k＜1/3
logk=-(ak+b) ---(1)
log2k=-(2ak+b) ---(2)
log3k=-(3ak+b) ---(3)
(1)-(2):
log(1/2)=ak
(2)+(3):
log(2/3)=ak
Which is impossible
If 1/3＜k＜1/2
logk=-(ak+b) ---(1)
log2k=-(2ak+b) ---(2)
log3k=3ak+b ---(3)
(1)-(2):
log(1/2)=ak
(2)+(3):
log(6k2)=ak
6k2=1/2
k=1/√12 (Rejected)
If 1/2＜k＜1
logk=-(ak+b) ---(1)
log2k=2ak+b ---(2)
log3k=3ak+b ---(3)
(1)+(2):
log(2k2)=ak
(3)-(2):
log(3/2)=ak
2k2=3/2
k=√3/2
If 1＜k
logk=ak+b ---(1)
log2k=2ak+b ---(2)
log3k=3ak+b ---(3)
(2)-(1):
log2=ak
(3)-(2):
log(3/2)=ak
which is impossible
k=√3/2
After checking a, b value, it does not fix value
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Raceleader - 訪客
由 Raceleader 於 星期五 四月 04, 2003 8:17 pm
Since we cannot find the REAL root of t, so this has no solution, right.
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Raceleader - 訪客
由 Raceleader 於 星期五 四月 04, 2003 8:18 pm
3個實根:
You should give out the exact value of t 1st, if no, how to determine a and b
You should give out the exact value of t 1st, if no, how to determine a and b
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Raceleader - 訪客
由 Raceleader 於 星期五 四月 04, 2003 8:20 pm
If the value of t cannot be determined, just know their ratio, assume t=1 can also ok.
log1=a+b
log2=2a+b
log3=3a+b
What is a and b?
log1=a+b
log2=2a+b
log3=3a+b
What is a and b?
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Raceleader - 訪客
