Question

The question

Answer 1

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

Answer 2

To determine the domain of x, let's look at the domains of sqrt(x) and log(x).

Answer 3

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

Answer 4

For sqrt(x), the domain is x>=0. For log(x), the domain is x>0.

Answer 5

Sorry,( 5x-x2)/ is larger than 1, that's group2

Answer 6

\(\log_{10}\dfrac{(5x-x^2)}{4} > 0\) right ?

Answer 7

The equivalence is okay, it can be zero

Answer 8

can log be zero too ?

Answer 9

Combining both domains, we should find x such that \[\frac{ 5x -x^2 }{ 4 } > 1\]

Answer 10

Yes

Answer 11

r u sure

Answer 12

Whenever the expression inside is 1

Answer 13

The domain of log(x) is x>0.

Answer 14

@math1234 gave you the conclusion

Answer 15

so i have to solve this right \(\large \dfrac{5x-x^2}{4}>0,\implies x\in(0,5)\)

Answer 16

The inequality value on the right should be strictly greater than 1.

Answer 17

how ?

Answer 18

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

Answer 19

If you didn't have the sqrt(x) on the outside, then you would be correct with setting 0 for the inequality

Answer 20

how do u know 0<x<1 is negative for log x

Answer 21

Remember how log(x) is defined. It returns the exponent of base 10 such that 10^(y) is equal to x.

Answer 22

Negative exponents give you values between 0 and 1.

Answer 23

ok

Answer 24

but i think it should be \(\large \dfrac{5x-x^2}{4}\geq 1\) since \(0<x<1\) and not \(0<x\leq 1\)

Answer 25

Yes, you are correct. It can be 1 too, since log(1) yields 0, and sqrt(0) is allowed.

Answer 26

Nice catch!

Answer 27

ok thnx