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mathmath333

  • one year ago

\(\large \color{black}{\begin{align}& \normalsize \text{Find the domain of the function} \hspace{.33em}\\~\\ & f(x)=\log_{10} [1-\log_{10} (x^2-5x+16)] \hspace{.33em}\\~\\ \end{align}}\)

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  1. anonymous
    • one year ago
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    First thing's first - do you know what the domain of log (a) is ?

  2. mathmath333
    • one year ago
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    a>0

  3. anonymous
    • one year ago
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    Precisely! So from that we can deduce so much that 1 - log (x^2-5x+16) has to be >0 for that whole thing to exist.

  4. anonymous
    • one year ago
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    From that we can deduce that log (x^2-5x+16) has to be <1 so that 1 - log (x^2-5x+16) >0

  5. mathmath333
    • one year ago
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    ok

  6. anonymous
    • one year ago
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    But at the same time, for log (x^2-5x+16) to exist, (x^2-5x+16) also has to be >0.

  7. Michele_Laino
    • one year ago
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    we have this condition, for the existence of the outer logarithm: \[\Large \begin{gathered} 1 - {\log _{10}}\left( {{x^2} - 5x + 16} \right) > 0 \hfill \\ \hfill \\ {\log _{10}}\left[ {\frac{{10}}{{{x^2} - 5x + 16}}} \right] > 0 \hfill \\ \hfill \\ \frac{{10}}{{{x^2} - 5x + 16}} > 1 \hfill \\ \end{gathered} \]

  8. Michele_Laino
    • one year ago
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    furthermore we have the subsequent condition for the existence of the inner logarithm: \[\Large {x^2} - 5x + 16 > 0\]

  9. Michele_Laino
    • one year ago
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    then the domain of our function is given by the solution of this system: \[\Large \left\{ \begin{gathered} \frac{{10}}{{{x^2} - 5x + 16}} > 1 \hfill \\ \hfill \\ {x^2} - 5x + 16 > 0 \hfill \\ \end{gathered} \right.\]

  10. anonymous
    • one year ago
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    It sounds a little wrapped up but it will make sense along the way. Let's work this one step at a time. For log (x^2-5x+16) to exist, (x^2-5x+16) also has to be >0. We solve the quadratic we find that (x^2-5x+16) has complex roots and that means that (x^2-5x+16) is always positive so we need not worry about it! Now, for that whole thing to exist, we've concluded that 1 - log (x^2-5x+16) >0 We "flip" the log over a little and we have that 1 > log (x^2-5x+16) Express 1 as log (10) ( log(10) is 1 ) and you have log (10) > log (x^2-5x+16) Which equates to 10 > (x^2-5x+16) Or 0 > (x^2-5x+6) We solve this new quadratic and we have that (x^2-5x+6)=(x-2)*(x-3). What that means is that (x^2-5x+6) will look something like this: |dw:1435346669499:dw|

  11. anonymous
    • one year ago
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    And to my guess, the quadratic itself will look something like this |dw:1435347000458:dw|

  12. anonymous
    • one year ago
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    Therefore, for 0>x^2-5x+6, the quadratic must be negative and that only happens for x between (-3,-2).

  13. mathmath333
    • one year ago
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    is the answer \(-3<x<-2\) or \(2<x<3\)

  14. anonymous
    • one year ago
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    . . .

  15. anonymous
    • one year ago
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    -3<x<-2

  16. anonymous
    • one year ago
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    Yeah, it's the same thing really, solve either one of them and you'll get there.

  17. mathmath333
    • one year ago
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    \(\Large \left\{ \begin{gathered} \frac{{10}}{{{x^2} - 5x + 16}} > 1 \hfill \\ \hfill \\ {x^2} - 5x + 16 > 0 \hfill \\ \end{gathered} \right.\) by evaluating this i got \(2<x<3\)

  18. Michele_Laino
    • one year ago
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    hint: this inequality: \[\large \Large \frac{{10}}{{{x^2} - 5x + 16}} > 1\] is equivalent to these systems: \[\left\{ \begin{gathered} 10 > {x^2} - 5x + 16 \hfill \\ \hfill \\ {x^2} - 5x + 16 > 0 \hfill \\ \end{gathered} \right.\quad \cup \quad \left\{ {\begin{array}{*{20}{c}} \begin{gathered} 10 < {x^2} - 5x + 16 \hfill \\ \hfill \\ \end{gathered} \\ {{x^2} - 5x + 16 < 0} \end{array}} \right.\]

  19. anonymous
    • one year ago
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    Simply taking one example from your answer. If 2<x<3, take for x=2.5 x^2-5x+16=6.25-12.5+16=9.75 10/9.75 is not greater than 1. Which means your evaluation is wrong.

  20. mathmath333
    • one year ago
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    but wolfram gave this http://www.wolframalpha.com/input/?i=+domain+f%28x%29%3D%5Clog_%7B10%7D+%5B1-%5Clog_%7B10%7D+%28x%5E2-5x%2B16%29%5D

  21. mathmath333
    • one year ago
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    r u sure that \(10/9.75\) is not greater than \(1\)

  22. anonymous
    • one year ago
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    Oh wait, damn.

  23. anonymous
    • one year ago
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    Ahahah.

  24. anonymous
    • one year ago
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    Yeah, it's 2<x<3. flutter, where did I go wrong ?

  25. anonymous
    • one year ago
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    Big medal for whoever finds the point at which I went wrong with my asumptions.

  26. anonymous
    • one year ago
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    Oh, got it.

  27. mathmath333
    • one year ago
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    everything is correct except u graphed wrong

  28. anonymous
    • one year ago
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    The roots of the quadratic were 2 and 3 not -2 and -3.

  29. anonymous
    • one year ago
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    Big blunder on my side, I apologize.

  30. mathmath333
    • one year ago
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    np thnks

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