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anonymous

  • 5 years ago

Solve for x exactly (no graphing) in (ln x)^4 = ln x^4. You will get two answers. Their sum is approximately A) 4.291 B) 4.691 C) 5.091 D) 5.491 E) 5.891

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  1. anonymous
    • 5 years ago
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    So if you just want a second opinion on your answer, what answer did you get?

  2. anonymous
    • 5 years ago
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    If you just want to get the answer, use wolframalpha. If you want to learn you'll need to put in some effort.

  3. anonymous
    • 5 years ago
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    Actually I have no idea how to solve this one. Im not sure what they want. Any ideas?

  4. anonymous
    • 5 years ago
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    Personally I would start with a substitution to make it easier to see what's going on

  5. anonymous
    • 5 years ago
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    For example, lets let k = ln(x)

  6. anonymous
    • 5 years ago
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    What would we have for our new version of the equation?

  7. anonymous
    • 5 years ago
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    (k)^4= k^4

  8. anonymous
    • 5 years ago
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    not quite.

  9. anonymous
    • 5 years ago
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    Thats basically the same thing on both sides of the equal sign

  10. anonymous
    • 5 years ago
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    It should be \( k^4 = ln(x^4)\)

  11. anonymous
    • 5 years ago
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    because in general for some value a, \[ln(a^4) \ne (ln\ a)^4\]

  12. anonymous
    • 5 years ago
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    But there is something we can do with the x^4 because of the properties of logarithms

  13. anonymous
    • 5 years ago
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    k^4= 4ln(x) ?

  14. anonymous
    • 5 years ago
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    Yes

  15. anonymous
    • 5 years ago
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    And since we said k = ln(x) ?

  16. anonymous
    • 5 years ago
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    k^4 = 4k?

  17. anonymous
    • 5 years ago
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    indeed

  18. anonymous
    • 5 years ago
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    So solve for k

  19. anonymous
    • 5 years ago
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    do I divide by 4 first or do I take the 4th root?

  20. anonymous
    • 5 years ago
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    I would move it over to the same side so it equals 0.

  21. anonymous
    • 5 years ago
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    then factor it

  22. anonymous
    • 5 years ago
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    Actually that might not work out well

  23. anonymous
    • 5 years ago
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    I mean it's correct, but there might be something easier

  24. anonymous
    • 5 years ago
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    hmmmm im trying to factor it and im at k(k^3-4)

  25. anonymous
    • 5 years ago
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    Right, so either k = 0 or k^3 = 4

  26. anonymous
    • 5 years ago
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    That'll work ok

  27. anonymous
    • 5 years ago
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    So now lets go back to our definition of k

  28. anonymous
    • 5 years ago
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    ln(x) = 0 and ln(x)^3=4

  29. anonymous
    • 5 years ago
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    ln(x) = 0 & ln(x) = (cube root)4 ?

  30. anonymous
    • 5 years ago
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    Yep, now raise e to the power of both sides to get rid of the ln

  31. anonymous
    • 5 years ago
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    x= (cuberoot 4) e = 4.31?

  32. anonymous
    • 5 years ago
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    Should be: \[x = e^0 \text{ or } x = e^\sqrt[3]{4}\]

  33. anonymous
    • 5 years ago
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    Because we had \[ln(x) = 0 \text{ or } ln(x) = \sqrt[3]{4}\] and \[ln(a) = b \iff e^b = a\]

  34. anonymous
    • 5 years ago
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    5.891 is the answer. That was a tuffy for me.

  35. anonymous
    • 5 years ago
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    Thanks

  36. anonymous
    • 5 years ago
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    Indeed. Takes a bit of thinking

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