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anonymous

  • 5 years ago

how do you solve f(x)=2^3x+5 and how do you find f^-1 9x) ?

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  1. shadowfiend
    • 5 years ago
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    Use the log, Luke ;) Seriously though, when you see a variable in a power, you can use logs to pull it out. Since you're solving, set f(x) = 0 and get: \[0 = 2^{3x} + 5\] Now try to see if you can get that into a log form that makes sense.

  2. anonymous
    • 5 years ago
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    what do i do after i get 2^3x = -5 ?

  3. anonymous
    • 5 years ago
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    ?

  4. shadowfiend
    • 5 years ago
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    Ok, so remember: \[a^b = c\] \[log_a c = b\]

  5. shadowfiend
    • 5 years ago
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    Can you apply that above?

  6. anonymous
    • 5 years ago
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    i'll try...

  7. shadowfiend
    • 5 years ago
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    Hm... Wait... Hehehe. I have detected a problem! That requires the log of a negative number, which does not exist.

  8. anonymous
    • 5 years ago
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    ?

  9. anonymous
    • 5 years ago
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    so what do i do then?

  10. shadowfiend
    • 5 years ago
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    Just to double check, the equation is: \[f(x)=2^{3x}+5\] Right?

  11. anonymous
    • 5 years ago
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    yeah

  12. anonymous
    • 5 years ago
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    is that it? shouldn't there be more problem solving?

  13. shadowfiend
    • 5 years ago
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    Can your answer have imaginary numbers?

  14. anonymous
    • 5 years ago
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    it polly should my math teacher loves i

  15. shadowfiend
    • 5 years ago
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    Hahaha. Well, the log of -5 is 0.698970004 + 1.36437635 i :p

  16. anonymous
    • 5 years ago
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    ow and wow OK

  17. shadowfiend
    • 5 years ago
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    Hehe. Yeah it's kind of nasty. If it were 2^(3x) - 5 instead of + 5 things would be a lot better.

  18. anonymous
    • 5 years ago
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    thanx though--- awesome response math wizard (:

  19. shadowfiend
    • 5 years ago
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    Oh that's not me. Amistre's not here at the moment but he's the wizard around here ;)

  20. anonymous
    • 5 years ago
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    oh, wait how is the second part done?

  21. anonymous
    • 5 years ago
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    f ^-1 (x)?

  22. shadowfiend
    • 5 years ago
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    Post that as a separate question. I'm a bit rusty on function inverses :)

  23. anonymous
    • 5 years ago
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    k

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