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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align}& \normalsize \text{find the domain.}\hspace{.33em}\\~\\ & y=\sqrt{\log_{10} \left(\dfrac{5x-x^2}{4}\right)} \end{align}}\)

  2. anonymous
    • one year ago
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    To determine the domain of x, let's look at the domains of sqrt(x) and log(x).

  3. anonymous
    • one year ago
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    5x-x2is larger than 0.....group 1 Log (5x-x2)/4 is larger than 1 ... Group2 Find the intersection now and you are done

  4. anonymous
    • one year ago
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    For sqrt(x), the domain is x>=0. For log(x), the domain is x>0.

  5. anonymous
    • one year ago
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    Sorry,( 5x-x2)/ is larger than 1, that's group2

  6. mathmath333
    • one year ago
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    \(\log_{10}\dfrac{(5x-x^2)}{4} > 0\) right ?

  7. anonymous
    • one year ago
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    The equivalence is okay, it can be zero

  8. mathmath333
    • one year ago
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    can log be zero too ?

  9. anonymous
    • one year ago
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    Combining both domains, we should find x such that \[\frac{ 5x -x^2 }{ 4 } > 1\]

  10. anonymous
    • one year ago
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    Yes

  11. mathmath333
    • one year ago
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    r u sure

  12. anonymous
    • one year ago
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    Whenever the expression inside is 1

  13. anonymous
    • one year ago
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    The domain of log(x) is x>0.

  14. anonymous
    • one year ago
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    @math1234 gave you the conclusion

  15. mathmath333
    • one year ago
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    so i have to solve this right \(\large \dfrac{5x-x^2}{4}>0,\implies x\in(0,5)\)

  16. anonymous
    • one year ago
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    The inequality value on the right should be strictly greater than 1.

  17. mathmath333
    • one year ago
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    how ?

  18. anonymous
    • one year ago
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    Evaluating log(x) for 0<x<1 is possible, but it yields a negative value, which is outside of the domain for sqrt(x).

  19. anonymous
    • one year ago
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    If you didn't have the sqrt(x) on the outside, then you would be correct with setting 0 for the inequality

  20. mathmath333
    • one year ago
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    how do u know 0<x<1 is negative for log x

  21. anonymous
    • one year ago
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    Remember how log(x) is defined. It returns the exponent of base 10 such that 10^(y) is equal to x.

  22. anonymous
    • one year ago
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    Negative exponents give you values between 0 and 1.

  23. mathmath333
    • one year ago
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    ok

  24. mathmath333
    • one year ago
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    but i think it should be \(\large \dfrac{5x-x^2}{4}\geq 1\) since \(0<x<1\) and not \(0<x\leq 1\)

  25. anonymous
    • one year ago
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    Yes, you are correct. It can be 1 too, since log(1) yields 0, and sqrt(0) is allowed.

  26. anonymous
    • one year ago
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    Nice catch!

  27. mathmath333
    • one year ago
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    ok thnx

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