mathmath333
  • mathmath333
The question
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align}& \normalsize \text{find the domain.}\hspace{.33em}\\~\\ & y=\sqrt{\log_{10} \left(\dfrac{5x-x^2}{4}\right)} \end{align}}\)
anonymous
  • anonymous
To determine the domain of x, let's look at the domains of sqrt(x) and log(x).
anonymous
  • anonymous
5x-x2is larger than 0.....group 1 Log (5x-x2)/4 is larger than 1 ... Group2 Find the intersection now and you are done

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More answers

anonymous
  • anonymous
For sqrt(x), the domain is x>=0. For log(x), the domain is x>0.
anonymous
  • anonymous
Sorry,( 5x-x2)/ is larger than 1, that's group2
mathmath333
  • mathmath333
\(\log_{10}\dfrac{(5x-x^2)}{4} > 0\) right ?
anonymous
  • anonymous
The equivalence is okay, it can be zero
mathmath333
  • mathmath333
can log be zero too ?
anonymous
  • anonymous
Combining both domains, we should find x such that \[\frac{ 5x -x^2 }{ 4 } > 1\]
anonymous
  • anonymous
Yes
mathmath333
  • mathmath333
r u sure
anonymous
  • anonymous
Whenever the expression inside is 1
anonymous
  • anonymous
The domain of log(x) is x>0.
anonymous
  • anonymous
@math1234 gave you the conclusion
mathmath333
  • mathmath333
so i have to solve this right \(\large \dfrac{5x-x^2}{4}>0,\implies x\in(0,5)\)
anonymous
  • anonymous
The inequality value on the right should be strictly greater than 1.
mathmath333
  • mathmath333
how ?
anonymous
  • anonymous
Evaluating log(x) for 0
anonymous
  • anonymous
If you didn't have the sqrt(x) on the outside, then you would be correct with setting 0 for the inequality
mathmath333
  • mathmath333
how do u know 0
anonymous
  • anonymous
Remember how log(x) is defined. It returns the exponent of base 10 such that 10^(y) is equal to x.
anonymous
  • anonymous
Negative exponents give you values between 0 and 1.
mathmath333
  • mathmath333
ok
mathmath333
  • mathmath333
but i think it should be \(\large \dfrac{5x-x^2}{4}\geq 1\) since \(0
anonymous
  • anonymous
Yes, you are correct. It can be 1 too, since log(1) yields 0, and sqrt(0) is allowed.
anonymous
  • anonymous
Nice catch!
mathmath333
  • mathmath333
ok thnx

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