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chika_tika3007 Group Title

2^2x -6.2^x +8 =0

  • one year ago
  • one year ago

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  1. ganeshie8 Group Title
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    say k = \(\large 2^x\)

    • one year ago
  2. ganeshie8 Group Title
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    \(\large 2^{2x} -6.2^x +8 =0\) \(\large (2^{x})^2 -6.2^x +8 =0\)

    • one year ago
  3. ganeshie8 Group Title
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    plugin k and you will see a quadratic in k solve k

    • one year ago
  4. chika_tika3007 Group Title
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    ok but how abaout the number 8???

    • one year ago
  5. ganeshie8 Group Title
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    wat do u get after putting \(2^x=k\) ?

    • one year ago
  6. ganeshie8 Group Title
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    we get this :- \(\large k^2-6k+8=0\)

    • one year ago
  7. ganeshie8 Group Title
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    this is just a quadratic equation you knw how to solve a quadratic equaiton ha ?

    • one year ago
  8. chika_tika3007 Group Title
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    yes

    • one year ago
  9. ganeshie8 Group Title
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    good, factor it and solve k first

    • one year ago
  10. chika_tika3007 Group Title
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    humm ok ok i got it k thanks

    • one year ago
  11. ganeshie8 Group Title
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    sure ?

    • one year ago
  12. ganeshie8 Group Title
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    \(\large (k-2)(k-4) = 0\) \(k = 2, k=4\)

    • one year ago
  13. chika_tika3007 Group Title
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    i get the answer are -2 and -4

    • one year ago
  14. chika_tika3007 Group Title
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    yes..

    • one year ago
  15. ganeshie8 Group Title
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    when k = 2, \(2^x = k\) \(2^x = 2^1\) \(x = 1\)

    • one year ago
  16. ganeshie8 Group Title
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    when k = 4, \(2^x = k\) \(2^x = 4\) \(2^x = 2^2\) \(x = 2\)

    • one year ago
  17. ganeshie8 Group Title
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    so the solutions are : 1, 2

    • one year ago
  18. chika_tika3007 Group Title
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    yup yup

    • one year ago
  19. ganeshie8 Group Title
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    cool :)

    • one year ago
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