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heathernelly

  • 3 years ago

;)))

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  1. jim_thompson5910
    • 3 years ago
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    Similar Problem: \[\Large \log_{3}(60) - \log_{3}(12)\] \[\Large \log_{3}\left(\frac{60}{12}\right)\] \[\Large \log_{3}(5)\] So \[\Large \log_{3}(60) - \log_{3}(12) = \log_{3}(5)\]

  2. heathernelly
    • 3 years ago
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    that's all i have to do?

  3. jim_thompson5910
    • 3 years ago
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    pretty much, but just use your numbers in place of mine

  4. heathernelly
    • 3 years ago
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    yeah, thanks! do you know how to do the other one?

  5. jim_thompson5910
    • 3 years ago
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    in general \[\Large \log_{b}(x) - \log_{b}(y) = \log_{b}\left(\frac{x}{y}\right)\]

  6. jim_thompson5910
    • 3 years ago
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    what do 16 and 64 have in common

  7. heathernelly
    • 3 years ago
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    they both have a 6 in them, both even #'s

  8. jim_thompson5910
    • 3 years ago
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    good, but in terms of factors

  9. heathernelly
    • 3 years ago
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    2^6? i don't know :S :/

  10. jim_thompson5910
    • 3 years ago
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    both are powers of 2 or powers of 4

  11. heathernelly
    • 3 years ago
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    ahh, yeah

  12. jim_thompson5910
    • 3 years ago
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    so 16 = 2^4 64 = 2^6 and 16 = 4^1 64 = 4^3

  13. heathernelly
    • 3 years ago
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    that's it?

  14. jim_thompson5910
    • 3 years ago
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    well you need to rewrite both sides to have the same base

  15. jim_thompson5910
    • 3 years ago
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    for instance, say you had 2^x = 16 you would rewrite 16 as 2^4 to get 2^x = 2^4 and you can see that x must be 4 (since both bases are 2)

  16. jim_thompson5910
    • 3 years ago
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    so you're using this idea to solve the second problem

  17. heathernelly
    • 3 years ago
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    ohhh, well that's pretty simple! thank youu

  18. heathernelly
    • 3 years ago
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    actually waait, it's not! haha.. i just got confused

  19. jim_thompson5910
    • 3 years ago
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    with what

  20. jim_thompson5910
    • 3 years ago
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    where are you stuck

  21. heathernelly
    • 3 years ago
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    and you can see that x must be 4 (since both bases are 2) ??

  22. jim_thompson5910
    • 3 years ago
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    well if b^x = b^y, then x = y

  23. heathernelly
    • 3 years ago
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    i'm confused because you did 2 different problems..:/

  24. jim_thompson5910
    • 3 years ago
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    that's just an example

  25. heathernelly
    • 3 years ago
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    okay, that makes sense .. but i'm still confused on what to do after what you said ''16 = 2^4 64 = 2^6 and 16 = 4^1 64 = 4^3''

  26. jim_thompson5910
    • 3 years ago
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    oh i meant to say 16 = 4^2, my bad

  27. jim_thompson5910
    • 3 years ago
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    1/16=64^(4x-3) 1/(4^2) = (4^3)^(4x-3) ... since 16 = 4^2 and 64 = 4^3 4^(-2) = 4^(3*(4x-3)) so because the bases are now the same, we can say that the exponents must be the same so... -2 = 3*(4x-3) solve for x

  28. heathernelly
    • 3 years ago
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    ahh, got it :) thanks a lot!!!

  29. jim_thompson5910
    • 3 years ago
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    np

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