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ramul

  • 2 years ago

how to find x in 4^x+4^(1/x)=8

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  1. asnaseer
    • 2 years ago
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    is the 2nd term definitely \(4^{\frac{1}{x}}\) and not \(4^{-x}\)?

  2. ramul
    • 2 years ago
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    yes

  3. asnaseer
    • 2 years ago
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    have you been asked to use any specific methods? e.g. solve algebraically or numerically?

  4. ramul
    • 2 years ago
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    you can solve algebraically

  5. asnaseer
    • 2 years ago
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    I can see one answer just by observation. See if you can also spot it if you rearrange your equation as:\[4^x+4^{\frac{1}{x}}=4+4\]

  6. ramul
    • 2 years ago
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    then what will be the next step?

  7. asnaseer
    • 2 years ago
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    observe the equation carefully - the answer should just "jump out" :)

  8. ramul
    • 2 years ago
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    on which basis rearrangement is done? can it be done to solve the equation?

  9. asnaseer
    • 2 years ago
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    try and equate the \(4^x\) with the first 4 on the right-hand-side. Similarly try and equate the \(4^{1/x}\) with the second 4 on the right hand side. What value of x satisfies both?

  10. ramul
    • 2 years ago
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    actually is this legal process? is any other method is there?

  11. asnaseer
    • 2 years ago
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    :) I'm not sure if it is a "legal" mathematical proof or not, but that is the method I used. Maybe others can come up with alternatives?

  12. ramul
    • 2 years ago
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    can you ask this to any of your teacher (maths)?

  13. asnaseer
    • 2 years ago
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    I am not in school - I do maths as a hobby so I have no teacher as such :)

  14. asnaseer
    • 2 years ago
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    I left school many many years ago

  15. ramul
    • 2 years ago
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    on which level are you?

  16. asnaseer
    • 2 years ago
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    I studied Aeronautical Engineering when I was at University. I now work as a software engineer

  17. ramul
    • 2 years ago
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    i want to ask next question how to prove null set is subset of every set ?

  18. asnaseer
    • 2 years ago
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    <--- you should ask each question separately in the list to the left. you can do this by first closing this question.

  19. ramul
    • 2 years ago
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    done now can you help me?

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