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DarkBlueChocobo

  • one year ago

Help with logarithims

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  1. DarkBlueChocobo
    • one year ago
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  2. DarkBlueChocobo
    • one year ago
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    For the first one would you just fill in x for 0?

  3. DarkBlueChocobo
    • one year ago
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    to which we would get log(10)10+3

  4. freckles
    • one year ago
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    yes replace x with 0

  5. freckles
    • one year ago
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    \[\log_a(a)=1 \text{ where } a \in (0,1) \cup (1,\infty)\]

  6. freckles
    • one year ago
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    since a^1 is a

  7. DarkBlueChocobo
    • one year ago
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    I am confused are you saying log10(10)=1?

  8. freckles
    • one year ago
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    yes \[10 \in (1,\infty) \\ \text{ so } 10 \in (0,1) \cup (1,\infty) \\ \text{ so yes } \log_{10}(10)=1 \]

  9. DarkBlueChocobo
    • one year ago
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    so like then we just add 3? so f(0)=4?

  10. freckles
    • one year ago
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    since 10^1=10

  11. freckles
    • one year ago
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    yep

  12. DarkBlueChocobo
    • one year ago
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    and for the second one would you just divide log(10)20/log(10)3

  13. freckles
    • one year ago
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    yes

  14. DarkBlueChocobo
    • one year ago
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    so log(3)20=x so you just seperate? and then they both have the common log(10)? im trying to udnerstand why

  15. freckles
    • one year ago
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    you can solve the equation by first taking log_10 of both sides

  16. freckles
    • one year ago
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    \[\log_{10}(20)=\log_{10}(3^x) \\ \log_{10}(20)=x \log_{10}(3) \text{ by power rule } \]

  17. freckles
    • one year ago
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    then divide both sides be log_10(3)

  18. DarkBlueChocobo
    • one year ago
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    Alrights Thank you for clearing that up. Egh last night of cramming for this test

  19. freckles
    • one year ago
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    \[\frac{\log_{10}(20)}{\log_{10}(3)}=\frac{x \log_{10}(3)}{\log_{10}(3)} \\ \frac{\log_{10}(20)}{\log_{10}(3)}=x\]

  20. freckles
    • one year ago
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    you could have chosen any log base as long as that base is a number between 0 and 1 exclusive or between 1 and infinity exclusive

  21. freckles
    • one year ago
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    maybe they want you to express in base 5 log well take log_5( ) of both side instead

  22. freckles
    • one year ago
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    just doing an example... \[20=3^x \\ \log_5(20)=\log_5(3^x) \\ \log_5(20)=x \log_5(3) \\ \frac{\log_5(20)}{\log_5(3)}=x\]

  23. freckles
    • one year ago
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    notice that answer could be expressed as: \[x=\frac{\log_a(20)}{\log_a(3)} \text{ where } a \in (0,1) \cup (1,\infty)\]

  24. anonymous
    • one year ago
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    can we say 3^x=20?

  25. freckles
    • one year ago
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    yes equality is symmetrical if a=b then b=a

  26. anonymous
    • one year ago
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    the can we say x=log3(20)?

  27. freckles
    • one year ago
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    yes my answer will also conclude that if you put a as 3 \[x=\frac{\log_3(20)}{\log_3(3)}=\frac{\log_3(20)}{1}=\log_3(20)\]

  28. anonymous
    • one year ago
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    can't we say x as a decimal?

  29. freckles
    • one year ago
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    you mean approximate x ?

  30. anonymous
    • one year ago
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    yep

  31. anonymous
    • one year ago
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    3^1=3 3^2=9 3^3=27 so x must be between 2 and three.... and so on?

  32. anonymous
    • one year ago
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    But if we are asked to find x for 4 decimal places can we find it? :(

  33. freckles
    • one year ago
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    calculator would probably be your friend

  34. anonymous
    • one year ago
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    can we find it without using it?

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