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TrojanPoem
 one year ago
if 6^ (1/a) = 3^(1/b) = 2^ (1/g)
TrojanPoem
 one year ago
if 6^ (1/a) = 3^(1/b) = 2^ (1/g)

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TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1Prove that dw:1434806977129:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.2is that the determinant

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1Yeah , no expansion

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3what do we know about a,b,g are they integers

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.16^ (1/a) = 3^(1/b) = 2^ (1/g)

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1I thought of adding logs 6^ (1/a) = 3^(1/b) = 2^ (1/g) and get out a/b , b/g , g/a however failed

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ya i was about to say i think we need to get simpler expressions with logs

dan815
 one year ago
Best ResponseYou've already chosen the best response.2probably mess around with that more

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1dw:1434807544755:dw

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1dw:1434807600411:dw

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1dw:1434807629601:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh how about doing trace stuff

dan815
 one year ago
Best ResponseYou've already chosen the best response.2or eigen value stuff, to show this is dependant

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1Show out your work !

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i think we can get those expressions for a b g, by solving for eigen values given your matrix,

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3how is finding eigenvalues any easier than row reduction followed by expansion

dan815
 one year ago
Best ResponseYou've already chosen the best response.2that involves expansion right in there

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1what is eigenvalues ?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1You can easily get its value without expanding here you go

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3it is a fancy term whihch linear algebra freaks use a lot ;p

dan815
 one year ago
Best ResponseYou've already chosen the best response.2did u try t work with this expression

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, here is the result A/B = Log6/ Log 3

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1I even added them to the det

dan815
 one year ago
Best ResponseYou've already chosen the best response.2forget the divisions for now, how about the additions can u get simp exp for that

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1You don't want them divided ? maybe you can right log3/log 6 as log3_6

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1Hmm C/in2 + D/in3 how did you separated it into two ? C/ in2 + in3 + D/in2 + in3

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i didnt do it, but we can do it, just look at the possibilities, that is partial fractions

dan815
 one year ago
Best ResponseYou've already chosen the best response.2a/(e+f) = c/e + d/f there is always a way to find htis

dan815
 one year ago
Best ResponseYou've already chosen the best response.2where c and d are in terms of a

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1we have to find the constants c, d right ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ii dont see where this goes though .

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3It is easy to show that \(a+b+g=0\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434808938502:dw

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.1B = GLog3 / Log2 A = GLo6/Log2 A + b +g = Glo3/log2  GLog6/Log2 + g = GLog3  Glo3  Glo2 + Glo2 / Log2 = 0/Log2 = 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Same thing but below looks less messier \( 6^{1/a} = 3^{1/b} = 2^{1/g} = k\) \(a = \frac{\ln 6}{k}\) \(b = \frac{\ln 3}{k}\) \(g = \frac{\ln 2}{k}\) \(a+b+g = 0\)

dan815
 one year ago
Best ResponseYou've already chosen the best response.2a b and g are not = 0 so

dan815
 one year ago
Best ResponseYou've already chosen the best response.2we can safely factor out a+b+g out off everything

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3looks we're done! aren't we ?
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