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coolaidd Group Title

(1) Write log9+1/3log729 as a single logarithm. (2) Write log[2]5-1/2log[2]169 as a single logarithm.

  • one year ago
  • one year ago

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  1. jbovey Group Title
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    1) the 1/3 would just be the exponent of log(729) so log(729)^1/3 and then we use log rule log(ab)= log(a)+log(b) to get log(9)(729)^(1/3)

    • one year ago
  2. jbovey Group Title
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    log(6561)^(1/3)

    • one year ago
  3. coolaidd Group Title
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    so i solve that? @jbovey i got In9/in10

    • one year ago
  4. coolaidd Group Title
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    is that for the #(2)?

    • one year ago
  5. jbovey Group Title
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    no thats the answer for #1. Since it says just write as single log you wouldn't have to solve any further

    • one year ago
  6. jbovey Group Title
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    it just wants you to use the log rules and combine

    • one year ago
  7. coolaidd Group Title
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    oooh ok.. so what about #2?

    • one year ago
  8. jbovey Group Title
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    Im kind of confused by the way you wrote it. Can u draw the equation out so I can see?

    • one year ago
  9. coolaidd Group Title
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    yupp! hold on..

    • one year ago
  10. jbovey Group Title
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    I know you'll have to use log rule 3 which is log(a)-log(b)= log(a)/(b)

    • one year ago
  11. coolaidd Group Title
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    |dw:1354512532548:dw|

    • one year ago
  12. jbovey Group Title
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    so are 5 and 169 the exponents?

    • one year ago
  13. coolaidd Group Title
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    no those are regular sized..

    • one year ago
  14. geoffb Group Title
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    The 2s are bases. She means \(\log_{2} 5 - \frac{1}{2} \log_{2} 169\)

    • one year ago
  15. coolaidd Group Title
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    25 and 338?

    • one year ago
  16. jbovey Group Title
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    ohhhh sorry about that, thanks @geoffb

    • one year ago
  17. geoffb Group Title
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    No problem. :)

    • one year ago
  18. jbovey Group Title
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    wait so wouldnt that make 5 and 169 the exponents? @geoffb

    • one year ago
  19. geoffb Group Title
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    No, it wouldn't.

    • one year ago
  20. geoffb Group Title
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    You would just need to maintain base 2. It's nice because both logs use base 2.

    • one year ago
  21. campbell_st Group Title
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    just a quick point on your first question... \[\frac{1}{3}\log(729) = \log(\sqrt[3]{729}) = \log(9)\] so then \[\log(9) + \frac{1}{3}\log(729) = \log(9) + \log(9) = \log(9 \times 9) = \log(81).. or... 2\log(9)\] just a suggestion... when compared to \[\log(\sqrt[3]{6561})\] not quite the same things...

    • one year ago
  22. geoffb Group Title
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    So, like you said, you could start by moving the 1/2 up as an exponent. $$\large \log_{2} 5 - \frac{1}{2} \log_{2} 169 = \log_{2} 5 - \log_{2} 169^{\frac{1}{2}} = \log_{2} (\frac{5}{13})$$

    • one year ago
  23. jbovey Group Title
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    Thanks bro @campbell_st I was close lol

    • one year ago
  24. coolaidd Group Title
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    ok..so would that be the final answer @geoffb ?

    • one year ago
  25. jbovey Group Title
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    yeah thats the answer @coolaidd

    • one year ago
  26. coolaidd Group Title
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    thanks!

    • one year ago
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