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travis

  • 4 years ago

(logx)^(logx)=? pfft didn't do log for quite a while :P

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  1. Mertsj
    • 4 years ago
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    \[\log a ^{n}=nloga\]

  2. travis
    • 4 years ago
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    not logx^(logx)

  3. myininaya
    • 4 years ago
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    \[\text{ Let } y=(\log(x))^{\log(x)} \] Now do log( ) of both sides. \[\log(y)=\log[(\log(x))^{\log(x)}]\] Using properties of log we can rewrite as \[\log(y)=\log(x)\cdot \log(\log(x))\] Again we can use another property to rewrite this as Now I will write this as an exponential equation where the base is 10 assuming the base is 10 here. \[10^{\log(y)}=10^{\log(x)\cdot \log(\log(x))}\] \[y=10^{\log(x) \cdot \log(\log(x))}\] => \[(\log(x))^{\log(x)}=10^{\log(x) \cdot \log(\log(x))}\] Both still pretty ugly, but I honestly don't know what you want to show here.

  4. travis
    • 4 years ago
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    oh no, i have to proof its = x

  5. myininaya
    • 4 years ago
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    what is the base?

  6. myininaya
    • 4 years ago
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    i assumed it was 10

  7. travis
    • 4 years ago
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    10 i think question didnt write

  8. travis
    • 4 years ago
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    if it was another base im'm sure it wouldn't affect the answer

  9. myininaya
    • 4 years ago
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    The base would just be some other number if not 10

  10. myininaya
    • 4 years ago
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    \[(\log(x))^{\log(x)}=a^{\log(x) \cdot \log(\log(x))} \] if the base was a

  11. travis
    • 4 years ago
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    Solve the equation (logx)^logx=x

  12. travis
    • 4 years ago
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    thats how the question says

  13. myininaya
    • 4 years ago
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    Oh we are solving an equation

  14. travis
    • 4 years ago
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    log(logx)^logx=logx (logx)log(logx)=logx I cant cancel log on both sides right?

  15. bahrom7893
    • 4 years ago
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    yea u can

  16. myininaya
    • 4 years ago
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    \[\log(x)(\log(\log(x))-1)=0\]

  17. myininaya
    • 4 years ago
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    subtract log(x) on both sides

  18. myininaya
    • 4 years ago
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    both terms have log(x) in common

  19. myininaya
    • 4 years ago
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    you factor

  20. travis
    • 4 years ago
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    log(logx)=1 log(logx)=log10 logx=10 x=100 right?

  21. myininaya
    • 4 years ago
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    what about x=10 for log(x)=0 since log(10)=1

  22. travis
    • 4 years ago
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    oh wait log 100 isnt 10

  23. travis
    • 4 years ago
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    x= 10^10!

  24. myininaya
    • 4 years ago
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    \[x=10^{10}\]

  25. myininaya
    • 4 years ago
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    you have two solutions

  26. travis
    • 4 years ago
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    whats the other solution?

  27. travis
    • 4 years ago
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    i took out logx so logx=0 right

  28. travis
    • 4 years ago
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    or logx=1

  29. myininaya
    • 4 years ago
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    \[(\log(x))^{\log(x)}=x\] D log( ) on both sides \[\log(x) \cdot \log(\log(x))=\log(x)\] Now subtract log(x) on both sides \[\log(x) \cdot \log(\log(x))-\log(x)=0\] Now factor the expression on the left side \[\log(x)(\log(\log(x))-1)=0\] Now set both factors =0 So we have \[\log(x)=0 \text{ or } \log(\log(x))-1=0\] So we should solve both equations \[x=10 \text{ or } \log(\log(x))=1\] \[x=10 \text{ or } 10^{\log(\log(x))}=10^1\] \[x=10 \text{ or } \log(x)=10\] \[x=10 \text{ or } 10^{\log(x)}=10^{10}\] \[x=10 \text{ or } x=10^{10}\]

  30. myininaya
    • 4 years ago
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    But if we check both solutions....

  31. travis
    • 4 years ago
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    I divided the log away instead of subtracting it,now i get where i was wrong, thanks a lot that was very kind of you!

  32. travis
    • 4 years ago
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    cancel the 10 right?

  33. Zarkon
    • 4 years ago
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    I think you ment to say x=1 though after checking one sees that is not a solution

  34. bahrom7893
    • 4 years ago
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    Travis you could have divided, you just made a mistake here: log(logx)=1 log(logx)=log10 logx=10 then x !=100

  35. bahrom7893
    • 4 years ago
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    10^log(x) = 10^10 x = 10^10

  36. myininaya
    • 4 years ago
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    oh yeah zarkon for some rason i solve log(x)=1

  37. myininaya
    • 4 years ago
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    reason*

  38. travis
    • 4 years ago
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    Yea I was too busy manipulating the log I didn't convert the log properly

  39. myininaya
    • 4 years ago
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    log(x)=0 x=1 for sure! :)

  40. travis
    • 4 years ago
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    yea so x =1 and x= 10^10

  41. Zarkon
    • 4 years ago
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    x=1 is only a solution if you define \[0^0=1\]

  42. myininaya
    • 4 years ago
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    right which i don't define it that way

  43. travis
    • 4 years ago
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    but if you sub x = 10^10 it works too

  44. myininaya
    • 4 years ago
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    i would say that is the only solution

  45. myininaya
    • 4 years ago
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    because after checking x=1 we get 0^0 like zarkon said

  46. travis
    • 4 years ago
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    So reject that solution right

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