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anonymous

  • one year ago

Graph y=log 3(x+2)-1. Then, find the domain/range

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  1. anonymous
    • one year ago
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    what does "solve" mean in this context?

  2. anonymous
    • one year ago
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    Whoops! Not solve. Just graph and find the domain/range

  3. jim_thompson5910
    • one year ago
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    I would use a graphing calculator like desmos https://www.desmos.com/calculator to graph this out. You will have to use the change of base formula to make sure the expression is typed in properly you will have to type in log(x+2)/log(3) - 1

  4. anonymous
    • one year ago
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    Ok thanks :) Would the domain be all real numbers?

  5. jim_thompson5910
    • one year ago
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    why all real numbers?

  6. jim_thompson5910
    • one year ago
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    you should have this graph https://www.desmos.com/calculator/fs1pnhr9hx

  7. jim_thompson5910
    • one year ago
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    the left side of the graph seems to hit a "wall" of some sort

  8. anonymous
    • one year ago
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    Ok, now I do. And it does seem like that

  9. jim_thompson5910
    • one year ago
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    it turns out that you cannot plug in 0 or negative numbers into a log function so x+2 has to be positive x+2 > 0 leads to x > ???

  10. anonymous
    • one year ago
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    x=-2 ? x+2=0

  11. jim_thompson5910
    • one year ago
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    yeah x = -2 is your vertical asymptote. The graph does NOT touch the vertical asymptote. It only gets closer and closer and closer.

  12. jim_thompson5910
    • one year ago
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    solving x+2 > 0 for x gives x > -2, which is the domain

  13. anonymous
    • one year ago
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    Alright, I see :) Now what about the range?

  14. jim_thompson5910
    • one year ago
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    the left side shoots off to negative infinity (very quickly) the right side slowly approaches positive infinity. it just keeps growing forever

  15. anonymous
    • one year ago
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    so the range would be all real #'s or infinity

  16. jim_thompson5910
    • one year ago
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    all real numbers I think you're thinking of this notation \(\Large (-\infty,\infty)\)

  17. anonymous
    • one year ago
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    Yes :)

  18. anonymous
    • one year ago
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    Thanks!

  19. jim_thompson5910
    • one year ago
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    you're welcome

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