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anonymous

  • one year ago

What is the value of x in the equation.... (logarithms), equation will be posted in comments

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  1. anonymous
    • one year ago
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    \[\log_{8}x + \log_{4}x = 10/3 \]

  2. kc_kennylau
    • one year ago
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    Firstly, the law: \[\Large\log_ab=\dfrac{\log_nb}{\log_na}\] Substitute \(\Large a=8\), \(\Large b=x\) to get: \[\Large\log_8x=\dfrac{\log_nx}{\log_n8}\]

  3. kc_kennylau
    • one year ago
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    Substitute \(\Large a = 4\), \(\Large b = x\) to get: \[\Large\log_4x=\dfrac{\log_nx}{\log_n4}\]

  4. kc_kennylau
    • one year ago
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    What would be a reasonable \(\Large n\) to fit in?

  5. anonymous
    • one year ago
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    10?

  6. kc_kennylau
    • one year ago
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    \[\Large \log_8x+\log_4x=\dfrac{10}3\] \[\Large \dfrac{\log_nx}{\log_n8}+\dfrac{\log_nx}{\log_n4}=\dfrac{10}3\] I would actually choose \(\Large n = 2\) because \(\Large 4=2^2\) and \(\Large 8=2^3\)

  7. kc_kennylau
    • one year ago
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    Let's use \(\Large n = 10\) anyway.

  8. kc_kennylau
    • one year ago
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    \[\Large \dfrac{\log_{10}x}{\log_{10}2^3}+\dfrac{\log_{10}x}{\log_{10}2^2}=\dfrac{10}3\]

  9. anonymous
    • one year ago
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    I was debating whether to use 2, but I didn't know how to use the 2 to the power of 2 and 3 in the equation.

  10. kc_kennylau
    • one year ago
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    okay, let's use \(\Large n = 2\) then.

  11. kc_kennylau
    • one year ago
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    \[\Large \dfrac{\log_2x}{\log_22^3}+\dfrac{\log_2x}{\log_22^2}=\dfrac{10}3\]

  12. kc_kennylau
    • one year ago
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    \[\Large \dfrac{\log_2x}3+\dfrac{\log_2x}2=\dfrac{10}3\]

  13. kc_kennylau
    • one year ago
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    Are you able to find \(\Large \log_2x\) now?

  14. anonymous
    • one year ago
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    Sorry, I'm not completely sure how to finish it.

  15. anonymous
    • one year ago
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    Would we simplify them into one fraction?

  16. kc_kennylau
    • one year ago
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    Let's substitute \(\Large a \) as \(\Large \log_2x\).

  17. kc_kennylau
    • one year ago
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    The equation then becomes: \[\Large\frac a3+\frac a2=\frac{10}3\]

  18. anonymous
    • one year ago
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    a=4?

  19. kc_kennylau
    • one year ago
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    Yes :)

  20. anonymous
    • one year ago
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    x=16!

  21. kc_kennylau
    • one year ago
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    So \(\Large \log_2x=4\)

  22. kc_kennylau
    • one year ago
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    Yes :D

  23. kc_kennylau
    • one year ago
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    Smart boy/girl

  24. anonymous
    • one year ago
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    Haha, girl - Thank you so much!

  25. kc_kennylau
    • one year ago
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    no problem :pp

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