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katymichellexx Group Title

Solve the logarithmic equation. Be sure to reject and value of x that is not in the domain of the original logarithmic expression. Give the exact answer. log4(x+8)+log4(x-4)=3 (the 4 after the logs are subscript)

  • one year ago
  • one year ago

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  1. geoffb Group Title
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    What do you know about logs and what they *can't* be used to calculate?

    • one year ago
  2. katymichellexx Group Title
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    I'm confused about the part about rejecting values of x that are not in the domain.

    • one year ago
  3. katymichellexx Group Title
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    How do I know which value of x to use? When I take my x values and plug them back into the equation, more than one x value fits and makes everything come out correctly but I am unsure about which one to put down for answer.

    • one year ago
  4. katymichellexx Group Title
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    I am using MyMathLab so it's very sensitive, and I have to put in the exact correct answer

    • one year ago
  5. geoffb Group Title
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    Can you take a log of a negative number?

    • one year ago
  6. katymichellexx Group Title
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    No

    • one year ago
  7. geoffb Group Title
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    Good. And what about 0?

    • one year ago
  8. katymichellexx Group Title
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    No

    • one year ago
  9. geoffb Group Title
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    Perfect. So you know, then, that each of your logs have to be > 0. That's the limitation of your domain.

    • one year ago
  10. katymichellexx Group Title
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    I put in x=8 on my MathLab and got it right! You made it so easy for me, thank you!

    • one year ago
  11. geoffb Group Title
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    Yes, it is 8, but I want to make sure you know why.

    • one year ago
  12. katymichellexx Group Title
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    x^2-4x+8x-32=4^3 x^2+4x-32=64 -64 -64 x^2+4x-96=0 x-8=0 x+12=0 x=8 x=-12

    • one year ago
  13. geoffb Group Title
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    Oh, good! Nice job.

    • one year ago
  14. katymichellexx Group Title
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    (8)^2-4(8)-32=0 64-32-32=0 0=0

    • one year ago
  15. katymichellexx Group Title
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    Thanks. I know how to get there, just not which X to use

    • one year ago
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