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soccergal12

  • 3 years ago

solve: log(underscore:x)9 = 1/2

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  1. lgbasallote
    • 3 years ago
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    _ <--underscore (shift + minus sign)

  2. lgbasallote
    • 3 years ago
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    anyway... \[\log_x 9 = \frac 12\]

  3. lgbasallote
    • 3 years ago
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    right?

  4. soccergal12
    • 3 years ago
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    yes igbasallote

  5. soccergal12
    • 3 years ago
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    that's the equation

  6. lgbasallote
    • 3 years ago
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    \[\log_x 9 = \frac 12\] first step is to change it into exponential form \[x^{1/2} = 9\] do you agree with this form

  7. soccergal12
    • 3 years ago
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    yes

  8. lgbasallote
    • 3 years ago
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    good. so we need x. you have x^1/2 what do you think you should do to make that x^1/2 into just x^1?

  9. soccergal12
    • 3 years ago
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    could you log both sides, and bring down the 1/2

  10. lgbasallote
    • 3 years ago
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    hmm not exactly

  11. lgbasallote
    • 3 years ago
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    here's a hint \[\LARGE (a^3)^{1/3} = a^{3/3} = a^1 = a\] \[\LARGE (a^2)^{1/2} = a^{2/2} = a^1 = a\] \[\LARGE (a^{1/3})^3 = a^{3/3} = a^1 = a\]

  12. soccergal12
    • 3 years ago
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    oh, do you square both sides?

  13. lgbasallote
    • 3 years ago
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    yup

  14. soccergal12
    • 3 years ago
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    so x = 81?

  15. lgbasallote
    • 3 years ago
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    yup

  16. soccergal12
    • 3 years ago
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    perfect thanks!

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