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anonymous

  • 4 years ago

solve by creating same bases: 1/9 = 27 ^ x-1

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  1. Mertsj
    • 4 years ago
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    \[\frac{1}{9}=3^{-2}\]

  2. Mertsj
    • 4 years ago
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    \[27^{x-1}=(3^{3})^{x-1}\]

  3. Mertsj
    • 4 years ago
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    \[(3^{3})^{x-1}=3^{3x-3}\]

  4. Mertsj
    • 4 years ago
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    So:\[3^{-2}=3^{3x-3}\]

  5. Mertsj
    • 4 years ago
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    And -2=3x-3 1=3x 1/3=x

  6. anonymous
    • 4 years ago
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    Can you please help me with one more??? PLEASE!! begging you.

  7. Mertsj
    • 4 years ago
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    Post it.

  8. anonymous
    • 4 years ago
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    properties of logarithms: 3^x=5^x+2

  9. anonymous
    • 4 years ago
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    3^x=5^(x+2)

  10. Mertsj
    • 4 years ago
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    \[x \log_{10}3=(x+2)\log_{10}5 \]

  11. anonymous
    • 4 years ago
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    How did you get that?

  12. Mertsj
    • 4 years ago
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    Find the log of 3 and the log of 5 using your calculator and solve the equation.

  13. Mertsj
    • 4 years ago
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    \[a ^{n}=n \log_{10}a \]

  14. anonymous
    • 4 years ago
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    how did you know what the variables were?

  15. Mertsj
    • 4 years ago
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    That is the property of law of logs. Perhaps your book uses different variables. What the variables are doesn't matter. Did you book say: \[m ^{n}=n \log_{10}m \]

  16. anonymous
    • 4 years ago
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    oh eyes!

  17. anonymous
    • 4 years ago
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    yes

  18. Mertsj
    • 4 years ago
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    So use that property on 3^x

  19. Mertsj
    • 4 years ago
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    Would you not get xlog3?

  20. anonymous
    • 4 years ago
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    I think my answer is wrong :/

  21. Mertsj
    • 4 years ago
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    What did you get?

  22. anonymous
    • 4 years ago
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    4.52

  23. Mertsj
    • 4 years ago
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    Hang on.

  24. Mertsj
    • 4 years ago
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    xlog3=.477x and (x+2)log 5 = .699x+1.40

  25. Mertsj
    • 4 years ago
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    do you agree with that/

  26. anonymous
    • 4 years ago
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    Yes!

  27. Mertsj
    • 4 years ago
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    Subtract .699x from both sides.

  28. Mertsj
    • 4 years ago
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    -.222x=1.40

  29. anonymous
    • 4 years ago
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    Oh so the final answer is 1.40?

  30. Mertsj
    • 4 years ago
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    Now divide both sides by -.222

  31. anonymous
    • 4 years ago
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    Oh so -6.31?

  32. Mertsj
    • 4 years ago
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    yes.

  33. anonymous
    • 4 years ago
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    THANK YOU SO MUCH! seriously, you are my life-saver. Now I have one more, I don't want to make you have to show me but could I just ask you a question about it?

  34. Mertsj
    • 4 years ago
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    Yes

  35. anonymous
    • 4 years ago
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    http://gyazo.com/8020e5d281ab10e0d5802e2690ad9794

  36. anonymous
    • 4 years ago
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    I don't get if there's acertain equation I am supposed to apply because it looks so confusing.

  37. Mertsj
    • 4 years ago
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    This involves two laws of logs. On the left side you have the difference of two logs which is equal to the log of the quotient.

  38. Mertsj
    • 4 years ago
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    \[\log_{3} \frac{x+1}{x}\]

  39. Mertsj
    • 4 years ago
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    The right side is the property we used in the last problem.

  40. Mertsj
    • 4 years ago
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    \[2\log_{3}2=\log_{3}2^{2} \]

  41. Mertsj
    • 4 years ago
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    So now, if the logs are equal, and the bases are the same, the arguments must be equal

  42. Mertsj
    • 4 years ago
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    \[\frac{x+1}{x}=2^{2}\]

  43. Mertsj
    • 4 years ago
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    solve that equation.

  44. Mertsj
    • 4 years ago
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    Cross multiply

  45. Mertsj
    • 4 years ago
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    4x=x+1

  46. Mertsj
    • 4 years ago
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    3x=1 x=1/3

  47. anonymous
    • 4 years ago
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    thank you so so so much!

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