mathmath333 one year ago The question

1. 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}}

2. anonymous

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

3. 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

4. anonymous

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

5. anonymous

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

6. mathmath333

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

7. anonymous

The equivalence is okay, it can be zero

8. mathmath333

can log be zero too ?

9. anonymous

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

10. anonymous

Yes

11. mathmath333

r u sure

12. anonymous

Whenever the expression inside is 1

13. anonymous

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

14. anonymous

@math1234 gave you the conclusion

15. mathmath333

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

16. anonymous

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

17. mathmath333

how ?

18. anonymous

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

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

20. mathmath333

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

21. anonymous

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

22. anonymous

Negative exponents give you values between 0 and 1.

23. mathmath333

ok

24. mathmath333

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

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

26. anonymous

Nice catch!

27. mathmath333

ok thnx