anonymous
  • anonymous
2^2x -6.2^x +8 =0
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ganeshie8
  • ganeshie8
say k = \(\large 2^x\)
ganeshie8
  • ganeshie8
\(\large 2^{2x} -6.2^x +8 =0\) \(\large (2^{x})^2 -6.2^x +8 =0\)
ganeshie8
  • ganeshie8
plugin k and you will see a quadratic in k solve k

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anonymous
  • anonymous
ok but how abaout the number 8???
ganeshie8
  • ganeshie8
wat do u get after putting \(2^x=k\) ?
ganeshie8
  • ganeshie8
we get this :- \(\large k^2-6k+8=0\)
ganeshie8
  • ganeshie8
this is just a quadratic equation you knw how to solve a quadratic equaiton ha ?
anonymous
  • anonymous
yes
ganeshie8
  • ganeshie8
good, factor it and solve k first
anonymous
  • anonymous
humm ok ok i got it k thanks
ganeshie8
  • ganeshie8
sure ?
ganeshie8
  • ganeshie8
\(\large (k-2)(k-4) = 0\) \(k = 2, k=4\)
anonymous
  • anonymous
i get the answer are -2 and -4
anonymous
  • anonymous
yes..
ganeshie8
  • ganeshie8
when k = 2, \(2^x = k\) \(2^x = 2^1\) \(x = 1\)
ganeshie8
  • ganeshie8
when k = 4, \(2^x = k\) \(2^x = 4\) \(2^x = 2^2\) \(x = 2\)
ganeshie8
  • ganeshie8
so the solutions are : 1, 2
anonymous
  • anonymous
yup yup
ganeshie8
  • ganeshie8
cool :)

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