YLL討論網

Raceleader · 由 **Raceleader** 於星期五四月 04, 2003 10:01 pm

I can get 1000 for proving that no real solution.

索洛西 · 由 **索洛西** 於星期五四月 04, 2003 10:04 pm

出題目的可以得500

Raceleader · 由 **Raceleader** 於星期五四月 04, 2003 10:04 pm

500 for you

Searchtruth · 由 **Searchtruth** 於星期五四月 04, 2003 10:08 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...

上面是我三天前算ㄉ...答案不是對ㄇ?
而且~t值是唯一ㄉ~不是ㄇ?
=>t 唯一解為 √(3)/2

Searchtruth · 由 **Searchtruth** 於星期五四月 04, 2003 10:26 pm

Raceleader:
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 (Rejected)
Why did you reject this k?
Is it wrong?
I can't find out it ~ Could you tell me?

Raceleader · 由 **Raceleader** 於星期五四月 04, 2003 10:27 pm

k=√3/2, can you check all 3 equations with a and b value?
I am finding, I will amend if can find out.

Searchtruth · 由 **Searchtruth** 於星期五四月 04, 2003 11:00 pm

It is too difficult for me to check them , because I can't use some software which can help me to check.(a and b are too complicated!!)
Then thank you anyway!^^"

Raceleader · 由 **Raceleader** 於星期五四月 04, 2003 11:02 pm

I think a,b are not unique for that value of k

Searchtruth · 由 **Searchtruth** 於星期五四月 04, 2003 11:08 pm

Really?
But I still stick on my thought...
Because I think my method is right...@@"

Searchtruth · 由 **Searchtruth** 於星期六四月 05, 2003 10:33 am

Have you check the answer?
How about ?

Searchtruth · 由 **Searchtruth** 於星期六四月 05, 2003 1:12 pm

Nobody wants to discuss this question.......>"<

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[數學]求a.b?