TrojanPoem
  • TrojanPoem
if 6^ (-1/a) = 3^(1/b) = 2^ (1/g)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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TrojanPoem
  • TrojanPoem
Prove that |dw:1434806977129:dw|
dan815
  • dan815
is that the determinant
TrojanPoem
  • TrojanPoem
Yeah , no expansion

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More answers

ganeshie8
  • ganeshie8
what do we know about a,b,g are they integers
TrojanPoem
  • TrojanPoem
6^ (-1/a) = 3^(1/b) = 2^ (1/g)
dan815
  • dan815
|dw:1434807323802:dw|
TrojanPoem
  • TrojanPoem
I 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
  • dan815
ya i was about to say i think we need to get simpler expressions with logs
dan815
  • dan815
probably mess around with that more
TrojanPoem
  • TrojanPoem
|dw:1434807544755:dw|
TrojanPoem
  • TrojanPoem
|dw:1434807600411:dw|
TrojanPoem
  • TrojanPoem
|dw:1434807629601:dw|
dan815
  • dan815
oh how about doing trace stuff
dan815
  • dan815
or eigen value stuff, to show this is dependant
TrojanPoem
  • TrojanPoem
Show out your work !
dan815
  • dan815
i think we can get those expressions for a b g, by solving for eigen values given your matrix,
dan815
  • dan815
|dw:1434807849696:dw|
ganeshie8
  • ganeshie8
how is finding eigenvalues any easier than row reduction followed by expansion
dan815
  • dan815
ok i guess not lol xD
dan815
  • dan815
that involves expansion right in there
TrojanPoem
  • TrojanPoem
what is eigenvalues ?
TrojanPoem
  • TrojanPoem
You can easily get its value without expanding here you go
ganeshie8
  • ganeshie8
it is a fancy term whihch linear algebra freaks use a lot ;p
dan815
  • dan815
lool
TrojanPoem
  • TrojanPoem
:D
dan815
  • dan815
did u try t work with this expression
dan815
  • dan815
|dw:1434808112422:dw|
TrojanPoem
  • TrojanPoem
Yeah, here is the result -A/B = Log6/ Log 3
TrojanPoem
  • TrojanPoem
Log3 / Log 2 = B/G
TrojanPoem
  • TrojanPoem
Log6/ Lo 2 = -A/G
TrojanPoem
  • TrojanPoem
I even added them to the det
TrojanPoem
  • TrojanPoem
Any ideas ?
dan815
  • dan815
forget the divisions for now, how about the additions can u get simp exp for that
dan815
  • dan815
|dw:1434808467873:dw|
TrojanPoem
  • TrojanPoem
You don't want them divided ? maybe you can right log3/log 6 as log3_6
dan815
  • dan815
|dw:1434808505436:dw|
TrojanPoem
  • TrojanPoem
Hmm C/in2 + D/in3 how did you separated it into two ? C/ in2 + in3 + D/in2 + in3
dan815
  • dan815
i didnt do it, but we can do it, just look at the possibilities, that is partial fractions
dan815
  • dan815
a/(e+f) = c/e + d/f there is always a way to find htis
TrojanPoem
  • TrojanPoem
Got it
dan815
  • dan815
where c and d are in terms of a
TrojanPoem
  • TrojanPoem
we have to find the constants c, d right ?
dan815
  • dan815
yea
dan815
  • dan815
ii dont see where this goes though -.-
ganeshie8
  • ganeshie8
It is easy to show that \(a+b+g=0\)
ganeshie8
  • ganeshie8
|dw:1434808938502:dw|
TrojanPoem
  • TrojanPoem
B = GLog3 / Log2 A = -GLo6/Log2 A + b +g = Glo3/log2 - GLog6/Log2 + g = GLog3 - Glo3 - Glo2 + Glo2 / Log2 = 0/Log2 = 0
TrojanPoem
  • TrojanPoem
right a+ b + g = 0
ganeshie8
  • ganeshie8
Looks good!
TrojanPoem
  • TrojanPoem
Now ideas ?
dan815
  • dan815
well its solved then
ganeshie8
  • ganeshie8
Same 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
  • dan815
a b and g are not = 0 so
dan815
  • dan815
we can safely factor out a+b+g out off everything
dan815
  • dan815
nice
TrojanPoem
  • TrojanPoem
Now ideas ?
ganeshie8
  • ganeshie8
looks we're done! aren't we ?
dan815
  • dan815
|dw:1434809493246:dw|
TrojanPoem
  • TrojanPoem
LOL, I understood

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