UN.Muzik Posted February 16, 2008 Report Share Posted February 16, 2008 a and b = 2did they have to be different numbers?yes they doFine a = 6 b = 82^6 = 648^2 = 64case closedwhy?Why what? You wanted values of a and b that are different and integers that satisfy the equation you mentionedthose numbers satisfy the equationUnless I'm missing the real question here? Quote Link to comment Share on other sites More sharing options...
Spawn Posted February 16, 2008 Report Share Posted February 16, 2008 I dont know why you bother answering this dribbling retards maths posts in the first place. Quote Link to comment Share on other sites More sharing options...
NORA BATTY Posted February 16, 2008 Author Report Share Posted February 16, 2008 a and b variables?b=sq. root of 2^aseveral values, could be displayed in a graph Lol ur such a wasteman. a and b = 2did they have to be different numbers?yes they doFine a = 6 b = 82^6 = 648^2 = 64case closedwhy?Why what? You wanted values of a and b that are different and integers that satisfy the equation you mentionedthose numbers satisfy the equationUnless I'm missing the real question here?well other than guessing how do you find the solutions and how do you know they are the only ones? Quote Link to comment Share on other sites More sharing options...
Bacardi_Up Posted February 16, 2008 Report Share Posted February 16, 2008 If I remember correctly you have to take logarithms of both sides so that will be:log2^a=log2^bThe logs are to the base 10. If you at all familar with the use of logarithms then you will no they can be rearranaged quite easily to equal the same thing.I think when you can rearrange them to:aLog2=bLog2You have to rearrange further but it has been 4 years and I can't remember to following steps. Quote Link to comment Share on other sites More sharing options...
NORA BATTY Posted February 16, 2008 Author Report Share Posted February 16, 2008 there are infinite solutions since in 2^a=b^2, 2 is prime so b=2^c and a=2c.so for even values a, there exist b=2^(a/2) such that 2^a=b^2. Quote Link to comment Share on other sites More sharing options...
NORA BATTY Posted February 16, 2008 Author Report Share Posted February 16, 2008 what about the solution of a^b=b^a for integers a and b? Quote Link to comment Share on other sites More sharing options...
Ali La Pointe Posted February 16, 2008 Report Share Posted February 16, 2008 Lol ur such a wasteman.AHA HA HA HAAAAAAAAAAAAA Quote Link to comment Share on other sites More sharing options...
Mist Posted February 16, 2008 Report Share Posted February 16, 2008 find solution to 2^a=b^2It's too early for this !*storms out* Quote Link to comment Share on other sites More sharing options...
Bacardi_Up Posted February 16, 2008 Report Share Posted February 16, 2008 there are infinite solutions since in 2^a=b^2, 2 is prime so b=2^c and a=2c.so for even values a, there exist b=2^(a/2) such that 2^a=b^2.llow it. You confused the f*ck out of me.... like I said it has been 3-4 years since doing A'level maths.I dunno if this is a joke and I have fallen for it like the proverbial bait but the logarithmic method is the usual approach to identities such as these.Open your maths book and learn the logarithmic identities. Quote Link to comment Share on other sites More sharing options...
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