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Audrae_World

  • 3 years ago

Use properties of logarithms to expand the loarithmic expressions as much as possile. log_4 7-2 1) 7log4 4 2) 4log4 7 3) -4log4 7 4) -16log 7

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  1. saifoo.khan
    • 3 years ago
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    7-2?

  2. Audrae_World
    • 3 years ago
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    7-^2

  3. saifoo.khan
    • 3 years ago
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    \[\log_4 7^{-2} \to \log_4 (\frac{1}{49})\to \frac{\log \frac{1}{49}}{\log 4}\]

  4. saifoo.khan
    • 3 years ago
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    are the options you wrote correct? @Audrae_World

  5. Audrae_World
    • 3 years ago
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    Yes, that's what my book says

  6. Calcmathlete
    • 3 years ago
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    Isn't it \[\log_{4} 1 - \log_{4} 7^{2}\]

  7. Eyad
    • 3 years ago
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    Its expand not solve @saifoo.khan

  8. saifoo.khan
    • 3 years ago
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    Yes! im stuck. :/

  9. Eyad
    • 3 years ago
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    @Audrae_World : Is it \[\log_{4} 7^2\]

  10. Audrae_World
    • 3 years ago
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    That's the closest I've seen so far, but the one of answer choices says: 4log_4 7^2

  11. Calcmathlete
    • 3 years ago
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    Are you sure the problem is \[\log_{4} 7^{-2}\] ?

  12. Audrae_World
    • 3 years ago
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    Yes

  13. Calcmathlete
    • 3 years ago
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    Oh. I got it. Hold on.

  14. Calcmathlete
    • 3 years ago
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    \[\log_{4}7^{-2} = -2(\log_{4}7) = -2(\log_{7}7/\log_{7}4) = -2(1/\log_{7}2^{2})\] \[-2(\log_{7}2^{-2}) = -2 * -2(\log_{7}2) = 4\log_{7}2\]

  15. Calcmathlete
    • 3 years ago
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    Wait. Darn it...

  16. Eyad
    • 3 years ago
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    LOL @Calcmathlete :whats that !??

  17. Calcmathlete
    • 3 years ago
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    lol. I got so close!!! But then I realized that it isn't one of the answers...

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