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anonymous

  • 5 years ago

what is a solution for the equation 2^n=10?

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  1. anonymous
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=2^n+%3D10

  2. anonymous
    • 5 years ago
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    Did you try taking the square root?

  3. anonymous
    • 5 years ago
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    Margaret?

  4. anonymous
    • 5 years ago
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    Am I missing something? Why would you take the square root... ?

  5. anonymous
    • 5 years ago
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    To get the value of n that yields 10.

  6. anonymous
    • 5 years ago
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    take the square root and then..? it doesnt work

  7. anonymous
    • 5 years ago
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    Ack! dislexia

  8. anonymous
    • 5 years ago
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    Don't worry, your answer was still more helpful than Mr. Wolfram's

  9. anonymous
    • 5 years ago
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    Take the log

  10. anonymous
    • 5 years ago
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    im confused

  11. anonymous
    • 5 years ago
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    I read it as \(n^2\) not \(2^n\) I need my exponents small and above, that flat representation plays hell with my dislexia.

  12. anonymous
    • 5 years ago
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    Take the log of both sides.

  13. anonymous
    • 5 years ago
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    Take the logarithm of both sides (ln or log for calc friendly) to knock that n down.

  14. amistre64
    • 5 years ago
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    log10 right?

  15. anonymous
    • 5 years ago
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    Log is much more friendly since it's a 10 on the other side

  16. anonymous
    • 5 years ago
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    But any log will do.

  17. anonymous
    • 5 years ago
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    would the answer be.. n=10/log2?

  18. anonymous
    • 5 years ago
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    No, you forgot to take the log of 10 also.

  19. anonymous
    • 5 years ago
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    soo.. n=1/log2?

  20. anonymous
    • 5 years ago
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    I wish that in addition to medals they'd let us award slaps for people who give half-assed answers and wander off ;p

  21. anonymous
    • 5 years ago
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    Yes precisely!

  22. anonymous
    • 5 years ago
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    so i got it!!?? :)

  23. anonymous
    • 5 years ago
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    You did! =)

  24. anonymous
    • 5 years ago
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    thankyou for your help

  25. anonymous
    • 5 years ago
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    You're very welcome.

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