Please Help! given a+b=ab=a^b, probe a=b=2 Please help me on this: Given a+b=ab=a^b prove a=b=2 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
Okey doke, the (hopefully) non-erroneous method. a+b = ab => b = ab-a = a(b-1) ab = a^{b} => b = a^{b-1} so, b = a(b-1) = a^{b-1} dividing by a; b-1 = 1^{b-2} = 1 => b = 2 a+b = ab => a+2 = 2a => a=b=2
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, a^{b} = 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 a^{b}= ab = a+b 2^{2} = 2*2 = 2+2, as required. Thus, a=b=2.
Agreed. Not quite sure I follow here. If a = b(a-1), then it follows that a^{b} = ( b(a-1) )^{b} = b^{b}(a-1)^{b}
I figured where I went wrong. It was my dodgy handwriting... a=b(a-1) a^{b}=b^{b}(a-1)^{b} ab=b^{2}(a-1)=b^{b}(a-1)^{b} So, (a-1) = b^{b-2}(a-1)^{b} Then, 1 = b^{b-2}(a-1)^{b-1} This is true if b^{b-2} is 1, and (a-1)^{b-1} is 1. b^{b-2} = 1 if b-2=0, so b=2. (a-1)=1, so a=2.
I don't disagree, but that doesn't really prove that there are no other numbers a and b for which b^{b-2} and (a-1)^{b-1} are inverses. edit: I think the problem essentially boils down to prove no real solutions other than a=2 for: a^{a-2}=(a-1)^{a-1}
How about this? From (1):a^{b}=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: (b)*ln[(b/b-1)]=ln(b/(b-1))+ln(b) so ln(b/(b-1))^{b}=ln(b^{2}/(b-1)) which means that b^{b}=b^{2} therefore, b=2 and from a=b/(b-1) we get a=2/(2-1)=2
I think it's good up until here. The b should be in the brackets to make ln([b/(b-1)]^{b}) This problem is harder than it first seems 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.
Re: PLEASE HELP!! given a+b=ab=a^b, probe a=b=2 I. a+b=ab I. b=ab-a I. b=a(b-1) I. a=b/(b-1) II. ab=a^b 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) Using Bernoulli... (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 0=-1 No solution. Wow. Where's the flaw? I know the RHS must be zero. Hopital, maybe?
I think your flaw may be in making the transition above, I don't know Bernoulli's rule or how you changed the ln's into a limit, but the limits don't match up: lim(n->inf) [ ((b-1)^(1/n)-1) / (b^(1/n)-1) ] = ln(b-1)/ln(b) and 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.
Lonewolf, you are probably right and I might need to change my classes . But as far as I can tell , the result of both ln(b/(b-1))^{b} and what you wrote comes out the same, which is: ln(b^{b}/(b-1)^{b}))=ln(b^{2}/(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(b^{b}/(b-1)^{b}))=ln(b^{2}/(b-1) means this: b^{b}/(b-1)^{b}=b^{2}/(b-1). (By using e^{ln} definition). If b^{b}=b^{2} then b=2 and if (b-1)^{b}=(b-1)^{1} then b=1. however, we knew already that b cannot be 1.
Unfortunantly, for that last equation to be true, b^{b} does not need to equal b^{2} and (b-1)^{b} need not equal (b-1). As long as the ratio between the pairs is satisfied, the equation can be satisfied and so we haven't yet proved no other real numbers can satisfy the last equation.
My teacher gave me an elegant solution 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=a^{b-1} b-1 = b/a b=a^{b/a} b^{a}=a^{b} 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 = a^{2} a=b= 2 or 0 I think the proof would be more elegant if we can do it using pure algebra.
Re: My teacher gave me an elegant solution It is not! f(1/e) = log(1/e)/e = -log(e)/e = -1/e f(e) = log e / e = 1/e f(e^2) = log(e^2)/(e^2) = 2/e^2 =(2/e)*(1/e) < 1/e i.e., it goes up and then back down.
hey guys, i'm quite sure i solved the problem, but since i'm busy i'll try to post a formal proof by the end of the day.
First note that neither a nor b can be equal to zero or one: from "a+b=ab" 0 + b = 0*b -> b=0 -> a^{b} = 0^{0} (an undefined quantity) a + 0 = a*0 -> a=0 -> a^{b} = 0^{0} 1 + b = 1*b -> contradiction a + 1 = a*1 -> contradiction By establishing that a,b != {0,1}, we are now free to divide by {a, b, a-1, b-1} a*b = a + b a*b - a = b a*(b-1) = b a = b/(b-1) Let's pause for a moment and further realize that b cannot be less than zero: if (b<0) -> [a = -|b|/(-|b|-1)] -> (a>0) but this means a*b is negative and a^{b} is postive, which is a contradiction since they must equal each other. Continuing: a^{b} = a*b a^{b-1} = b ( b/(b-1) )^{b-1} = b (substitution from above) b^{b-2} = (b-1)^{b-1} |b^{b-2}| = |(b-1)^{b-1}| |b|^{b-2} = |b-1|^{b-1} ln|b|^{b-2} = ln|b-1|^{b-1} (b-2)ln|b| - (b-1)ln|b-1| = 0 We now set f(x) = (x-2)ln|x| - (x-2)ln|x-1|, x!={0,1}. The zeros of f(x) are the only possible solutions for b as f(b) must equal zero. Finding these zeros is a difficult task and requires analyzing the derivatives. f'(x) = ln|x/(x-1)| - 2/x, x!={0,1} f''(x) = (x-2)/( x^{2}(x-1) ), x!={0,1} Since f''(x) is the only function we really recognize, we us it to find out what f'(x) looks like and then we use that to find out what f(x) looks like. Code (Text): + + - + f''(x) <---------|---------|---------|---------> 0(und) 1(und) 2 Lim[ x-> +/-infinity, f'(x) ] = 0 f'(0.1) < 0 and f'(0.95) > 0 (just plug it in) f'(1.1) > 0 and f'(1.90) < 0 This means that f'(x) must have only two zero which exist on (0,1) and (1,2) (because f' is continous within the intervals inbetween) which we will label c_{1} and c_{2} respectively. This in turn implies that the sign chart of f'(x) looks as follows: Code (Text): + - + + - f'(x) <---------|---------|---------|---------|---------> 0(und) c1 1(und) c2(c2<2) Limit[ x->1, f(x) ] = 0 (though not defined at 1) f(-4) < 0 and f(-3) > 0 f(0.1) > 0 and f(0.9) < 0 f(1.1) > 0 and f(2.1) < 0 Finally, we use the last set of information to note that there are precisely three zeros of f(x). Since limit of f(x) as x approaches 1 is zero and f(x) is increasing on c_{1} to c_{2} there can't be a zero within this interval. However the other three intervals do have zeros because f(x) is continous on them and changes sign: (-inf, 0), (0, c_{1}), (c_{2}, +inf). We can label these zeros as b_{1}, b_{2}, and b_{3} respectively. b_{1}, b_{2}, and b_{3} are the only possible solutions for b because f(b) must equal zero as I stated above. But also remember that b > 0 so that b_{1}, which is in (-inf, 0), is ruled out. If we assume b_{2} is correct than a is negative: 0<b_{2}<c_{1}<1 -> a = |b|/(|b|-1) = pos/neg = neg. But this would mean that a^{b} has a negative base. This leads to a complex number for a^{b} which certainly cannot equal a+b. (Although if you simply take the neg base as meaning -|a|^{b}, then this actually would look like a solution as i tried plugging in numbers and it comes extremely close). Anyways, now we have eliminated all but one possible solution, b_{3}, which exists in the interval [c_{2}, +inf) where c_{2} is less than 2. By inspection we know f(2) = 0 and therefore b_{3} must equal 2. We now try setting b=2 and seeing if this really is a solution: a = 2/(2-1) = 2 2 +2 = 2^{2} = 2*2 = 4 I know this has to be the ugliest solution you would ever want to see, and i've omitted some of the stuff to keep this from being any longer... but i think this really proves that (a=2,b=2) is the only solution. Actually, like i said, b_{2} might work. Switch your calculator out of complex mode and into real mode and try it out: (a=-3.14104155643, b=.75851486)
a, b integers ? If both a and b are integers, then a must equal to b. a^{b}=b^{a} by the fundamental theorem of arithmetic, a = b On one hand, y=a^{b} where y is an integer and it can be uniquely factorized. (a is a prime number) On the other hand, y = b^{a} where b is a prime number. We can prove my contradition that a and b are even numbers. Therefore a=b=2 I'm sorry to cause confusion here. Yes, f(x) is strictly increasing when e>x>0 and it is strictly decreasing when x>e. It seems that we have infinitely many solutions if a and b aren't integers.