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mathmath333
 one year ago
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mathmath333
 one year ago
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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align}& \normalsize \text{find the domain.}\hspace{.33em}\\~\\ & y=\sqrt{\log_{10} \left(\dfrac{5xx^2}{4}\right)} \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0To determine the domain of x, let's look at the domains of sqrt(x) and log(x).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.05xx2is larger than 0.....group 1 Log (5xx2)/4 is larger than 1 ... Group2 Find the intersection now and you are done

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For sqrt(x), the domain is x>=0. For log(x), the domain is x>0.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry,( 5xx2)/ is larger than 1, that's group2

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\log_{10}\dfrac{(5xx^2)}{4} > 0\) right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The equivalence is okay, it can be zero

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0can log be zero too ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Combining both domains, we should find x such that \[\frac{ 5x x^2 }{ 4 } > 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Whenever the expression inside is 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The domain of log(x) is x>0.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@math1234 gave you the conclusion

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0so i have to solve this right \(\large \dfrac{5xx^2}{4}>0,\implies x\in(0,5)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The inequality value on the right should be strictly greater than 1.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Evaluating log(x) for 0<x<1 is possible, but it yields a negative value, which is outside of the domain for sqrt(x).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you didn't have the sqrt(x) on the outside, then you would be correct with setting 0 for the inequality

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0how do u know 0<x<1 is negative for log x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Remember how log(x) is defined. It returns the exponent of base 10 such that 10^(y) is equal to x.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Negative exponents give you values between 0 and 1.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0but i think it should be \(\large \dfrac{5xx^2}{4}\geq 1\) since \(0<x<1\) and not \(0<x\leq 1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, you are correct. It can be 1 too, since log(1) yields 0, and sqrt(0) is allowed.
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