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

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

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

Have you check the answer?
How about ?

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...@@"

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

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!^^"

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.

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?

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

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

500 for you

由 **索洛西** 於星期五四月 04, 2003 10:04 pm

出題目的可以得500

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

I can get 1000 for proving that no real solution.

由 **索洛西** 於星期五四月 04, 2003 9:57 pm

也許這個題目出的不好...

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?

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

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.

由 **索洛西** 於星期五四月 04, 2003 8:15 pm

因為題目只給三個實根的比值，所以t 的值並不唯一

由 **Raceleader** 於星期五四月 04, 2003 8:12 pm

what is the answer of t?

由 **索洛西** 於星期五四月 04, 2003 7:55 pm

This is answer:
設3個實根為t.2t.3t
由圖形與交點位置可知 0<t<1，2t>1 ， 3t>1
l logt l=-logt=at+b
l log2t l=log2+logt=a(2t)+b
l log3t l=log3+logt=a(3t)+b
解a=['根號3' 分之2 ]*log3/2
b=log4/3根號3

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

由 **Raceleader** 於星期五四月 04, 2003 7:27 pm

logk, log2k, log3k cannot be all positive or negative...

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[quote="Raceleader"]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?[/quote]