## anonymous one year ago solve the inequality? x^2e^xlnx>0

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1. anonymous

$x^2 e^x lnx >0$

2. freckles

you only need to solve ln(x)>0

3. anonymous

isn't x undefined then?

4. freckles

no... $\ln(x)=y \implies e^{y}=x \\ \text{ here we have } y=0 \\ \text{ so we have } e^{0}=x$ so how do you solve the equation ln(x)=0

5. freckles

like what is the solution

6. freckles

and remember y=ln(x) looks like this: |dw:1440878660198:dw|

7. anonymous

well e^0 is 1

8. freckles

drawing got cut off but that line is suppose to be y=0

9. freckles

so x>1

10. freckles

x^2e^xln(x)>0 when x>1

11. anonymous

so in interval notation, is it $(1,\infty)$

12. freckles

yep

13. anonymous

thank you so much

14. freckles

do you understand how we got that though ?

15. freckles

|dw:1440878804304:dw|

16. freckles

e^x was always positive x^2 was always positive ln(x) was going to determine whether or not e^x*x^2*ln(x) was positive or not

17. anonymous

yeah, I'm actually kind of mad at myself that I didnt change it to exponential notation.

18. freckles

just for fun x^2e^xln(x)<0 has solution (0,1)

19. freckles

I didn't include anything to the left of 0 because ln(x) is undefined there

20. freckles

left of 0 or 0*

21. dinamix

|dw:1440879090764:dw| your graph for function is like that so f(x)>0 in ] 1; +infinity [ @hailbug