|Apr24-03, 08:27 AM||#1|
Please Help! given a+b=ab=a^b, probe a=b=2
Please help me on this:
it seems like a very simple task but I have lost my sleep for the past
couple of nights over it.
please help or ill be [:((] [:((] [:((] [:((] [:((]
|Apr24-03, 06:52 PM||#2|
Sorry, you realised it before I did.
|Apr24-03, 07:11 PM||#3|
Okey doke, the (hopefully) non-erroneous method.
a+b = ab => b = ab-a = a(b-1)
ab = ab => b = ab-1
so, b = a(b-1) = ab-1
dividing by a;
b-1 = 1b-2 = 1 => b = 2
a+b = ab => a+2 = 2a => a=b=2
|Apr24-03, 07:25 PM||#4|
Please Help! given a+b=ab=a^b, probe a=b=2
b-1 = ab-2
|May24-03, 12:38 PM||#5|
This may well be a bit late, but I was bored and browsing through some old posts.
Using a+b = ab, we get a = ab-b = b(a-1)
So, ab = b(a-1)b
But, b*b(a-1)b = ab, so, a = b*(a-1)b
Now, b*(a-1)b = b(a-1), so (a-1)b = (a-1)
This is true if a=2, or b=1.
Substituting b=1 into ab = a+b, a*1 = a+1, which leads to 0=1, a contradiction, thus b is not 1.
Substituting a=2 into ab = a+b, 2b = b+2, so b=2.
Substituting these two values into ab= ab = a+b
22 = 2*2 = 2+2, as required. Thus, a=b=2.
|May24-03, 04:34 PM||#6|
If a = b(a-1), then it follows that ab = ( b(a-1) )b = bb(a-1)b
|May24-03, 04:42 PM||#7|
Quite right. Well spotted.
|May24-03, 05:06 PM||#8|
I figured where I went wrong. It was my dodgy handwriting...
So, (a-1) = bb-2(a-1)b
Then, 1 = bb-2(a-1)b-1
This is true if bb-2 is 1, and (a-1)b-1 is 1.
bb-2 = 1 if b-2=0, so b=2. (a-1)=1, so a=2.
|May24-03, 09:31 PM||#9|
I think the problem essentially boils down to prove no real solutions other than a=2 for:
|May25-03, 01:52 AM||#10|
How about this?
From (1):ab=ab we get (b)*ln(a)=ln(a)+ln(b) and
From (2):a+b=ab we get a=b/(b-1) so ln(a)=ln(b/(b-1))
subsitute (2) into (1) to get:
ln(b/(b-1))b=ln(b2/(b-1)) which means that
b=2 and from a=b/(b-1) we get a=2/(2-1)=2
|May25-03, 03:39 AM||#11|
This problem is harder than it first seems [s(]
There are other inverses to my equation after all...take b=1, then a could be any real number, except 1...oh well. I think Thoth's onto something though.
|May25-03, 04:55 PM||#12|
I into II. b^2/(b-1)=(b/(b-1))^b
2ln b - ln(b-1) = b ln b - b ln(b-1)
(2-b)ln b = (1-b)ln(b-1)
(2-b)/(1-b) = ln(b-1)/ln(b)
Use definition of ln...
(2-b)/(1-b) = lim(n->[oo]) n((b-1)^(1/n)-1) / n(b^(1/n)-1)
n cancels ...
(2-b)/(1-b) = lim(n->[oo]) ((b-1)^(1/n)-1) / (b^(1/n)-1)
(2-b)/(1-b) = lim(n->[oo]) (1 + b/n -1/n -1) / (1 + b/n -1)
(2-b)/(1-b) = lim(n->[oo]) (b-1)/b
(2-b)/(1-b) = (b-1)/b
2b-b^2 = 2b-b^2-1
Wow. Where's the flaw? I know the RHS must be zero.
|May25-03, 05:33 PM||#13|
lim(n->inf) [ ((b-1)^(1/n)-1) / (b^(1/n)-1) ] = ln(b-1)/ln(b)
lim(n->inf) [ (1 + b/n -1/n -1) / (1 + b/n -1) ] = (b-1)/b
I would be a little hesitant to use ln's anyway because you don't know whether the quantities are negative. But it looks like a new avenue.
|May26-03, 01:21 AM||#14|
Lonewolf, you are probably right and I might need to change my classes[a)]. But as far as I can tell[8)], the result of both ln(b/(b-1))b and what you wrote comes out the same, which is: ln(bb/(b-1)b))=ln(b2/(b-1). However I give your opinion the benefit of the doubt since typing mathematics does not come easy for me[:((].
A major factor is to look at a=b/(b-1) and ln(a)=ln(b/(b-1)) and ask what values for b make sense for a?
Here a=b/(b-1) first we notice that b cannot be =1 because of the singularity at b=1.
After noticing this we then have to ask how about if b>1 and b<1 what would happen then. As b increases in value how a is being effected. We also need to know what values for both a and b satisfies the given conditions in the problem.
In ln(bb/(b-1)b))=ln(b2/(b-1) means this: bb/(b-1)b=b2/(b-1). (By using eln definition). If bb=b2 then b=2 and if (b-1)b=(b-1)1 then b=1. however, we knew already that b cannot be 1. [:D]
|May26-03, 09:57 AM||#15|
|May28-03, 02:37 AM||#16|
This question has troubled me for some time, finally I gave it to my math teacher and he gave me an elegant solution today.
b-1 = b/a
a log b = b log a
log a / a = log b / b
a=b (as the function f(x)=logx/x is strictly increasing)
=>2a = a2
a=b= 2 or 0
I think the proof would be more elegant if we can do it using pure algebra.
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