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chika_tika3007

  • one year ago

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

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  1. ganeshie8
    • one year ago
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    say k = \(\large 2^x\)

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

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

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

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

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

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

  8. chika_tika3007
    • one year ago
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    yes

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

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

  11. ganeshie8
    • one year ago
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    sure ?

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

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

  14. chika_tika3007
    • one year ago
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    yes..

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

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

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

  18. chika_tika3007
    • one year ago
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    yup yup

  19. ganeshie8
    • one year ago
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    cool :)

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