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anonymous

  • 5 years ago

Exponential Equations help? Determine the exact value of x: (1/8)^(x-3) = 2 * 16^(2x+1)

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  1. anonymous
    • 5 years ago
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    Are you using logarithms to solve?

  2. anonymous
    • 5 years ago
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    can it be solved without logarithms??

  3. anonymous
    • 5 years ago
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    I don't know how to solve w/o logs

  4. anonymous
    • 5 years ago
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    okay...logarithms are fine then :)

  5. anonymous
    • 5 years ago
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    Haha :) First, take the log of both sides

  6. anonymous
    • 5 years ago
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    Then apply your log rules...in other words......

  7. anonymous
    • 5 years ago
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    \[\log (1/8)^{x-3}= (x-3) \log (1/8)\]

  8. anonymous
    • 5 years ago
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    and \[\log (2 * 16^{2x +1}) = \log 2 + \log 16^{2x+1}\]

  9. anonymous
    • 5 years ago
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    this equals \[\log 2 + (2x+1) \log 16\]

  10. anonymous
    • 5 years ago
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    so now you have\[(x-3) \log (1/8)= \log 2 + (2x + 1) \log 16\]

  11. anonymous
    • 5 years ago
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    evaluate the logs and solve the equation for x

  12. anonymous
    • 5 years ago
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    There may be another way to solve, but I'm not sure

  13. anonymous
    • 5 years ago
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    thanks :) ..but can you explain what you mean by evaluate the logs? i've never done them before, really. :S

  14. anonymous
    • 5 years ago
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    ok, forget all that....i just figured out a better way :)

  15. anonymous
    • 5 years ago
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    lol okay

  16. anonymous
    • 5 years ago
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    1/8 = \[2^{-3}\]

  17. anonymous
    • 5 years ago
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    and 16 = \[2^{4}\]

  18. anonymous
    • 5 years ago
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    This gives you \[2^{-3(x-3)}= 2^{1}*2^{4(2x+1)}\]

  19. anonymous
    • 5 years ago
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    For the exponents, use distributive property

  20. anonymous
    • 5 years ago
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    \[2^{-3x+9}= 2^{1}*2^{8x +1}\]

  21. anonymous
    • 5 years ago
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    on the right side, you can combine those by adding expontents

  22. anonymous
    • 5 years ago
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    \[2^{-3x+9}=2^{8x+2}\]

  23. anonymous
    • 5 years ago
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    Now, since the bases are equal, this means the exonents are =, so\[-3x + 9 = 8x + 2\]

  24. anonymous
    • 5 years ago
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    Then, solve the equation for x

  25. anonymous
    • 5 years ago
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    haha....that was easier than logs :)

  26. anonymous
    • 5 years ago
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    yay! this makes so much sense to me. thanks so much :)

  27. anonymous
    • 5 years ago
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    You're very welcome!

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