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anonymous

  • 5 years ago

solve for x: log$2(x+1)+log$2(3x-5)=log$2(5x-3)+2

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    (x+1)(3x-5) = 4(5x-3)

  3. anonymous
    • 5 years ago
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    okay......can you keep going?

  4. anonymous
    • 5 years ago
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    3x^2 -2x -5 = 20x -12 3x^2 - 22x +7 = 0

  5. anonymous
    • 5 years ago
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    does the last line factor into (x-7)(3x-1)?

  6. anonymous
    • 5 years ago
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    yes

  7. anonymous
    • 5 years ago
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    x = 7 or 1/3 now substitute both these values into ur initial eqn to double check

  8. anonymous
    • 5 years ago
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    okay and on the first line you wrote 4(5x-3).....how did you get the four? 2^2?

  9. anonymous
    • 5 years ago
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    yes 2 can be written as log 4 to base 2

  10. anonymous
    • 5 years ago
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    and then we use log a + log b =log (ab)

  11. anonymous
    • 5 years ago
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    thanks you. thanks were i kept going wrong. The solutions both work. :)

  12. anonymous
    • 5 years ago
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    no prob miss

  13. anonymous
    • 5 years ago
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    can i ask one more thing? if you are solving for x and you take the square root to isolate x, do you include both the positive and negative answers?

  14. anonymous
    • 5 years ago
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    yes

  15. anonymous
    • 5 years ago
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    like lets say x^2 = k then x = k AND -k

  16. anonymous
    • 5 years ago
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    in some special situations we can neglect the negative answer..bt generally it is essential to consider both

  17. anonymous
    • 5 years ago
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    thanks you!

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