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Grazes

  • 3 years ago

logx^3=(logx)^2 solve for x

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  1. Kanwar245
    • 3 years ago
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    3logx = (log x)^2 log x (log x - 3) = 0 log x = 0 or log x = 3 x = 1 or e^3

  2. some_someone
    • 3 years ago
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    yeah @Kanwar245 has it :)

  3. Grazes
    • 3 years ago
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    I meant|dw:1365283288597:dw|

  4. Grazes
    • 3 years ago
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    Same thing?

  5. Kanwar245
    • 3 years ago
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    yes I used log rules and then factored log x

  6. some_someone
    • 3 years ago
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    actually @Kanwar245 wouldn't x = 1 and x = e

  7. Kanwar245
    • 3 years ago
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    log e^3 = 3

  8. Kanwar245
    • 3 years ago
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    log 1 = 0

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