## animalsavior94 4 years ago condense this to a single quantity: 1/2lnX +ln(x+3)-ln(x^2 +1) thanks.

1. Owlfred

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

2. LagrangeSon678

This appears to become a quadratic equation

3. animalsavior94

really? well im not sure because its logarithms and u need to condense it.

4. animalsavior94

how did you get that?

5. satellite73

well it is wrong, so don't write it!

6. satellite73

7. animalsavior94

oh i see could you please explain it to me?

8. satellite73

$\frac{1}{2}\ln(x)=\ln(\sqrt{x})$

9. satellite73

because $\ln(x^n)=n\ln(x)$

10. satellite73

here i am going from the right hand side to the left hand side.

11. satellite73

$\frac{1}{2}\ln(x)=\ln(x^{\frac{1}{2}})=\ln(\sqrt{x})$

12. satellite73

then we use $\ln(ab) = \ln(a)+ln(b)$ again from the right to the left.

13. satellite73

so we get $\ln(\sqrt{x})+\ln(x+3)=\ln(\sqrt{x}(x+3))$

14. satellite73

then we use $\ln(a)-\ln(b)=\ln(\frac{a}{b})$

15. satellite73

to get $\ln(\sqrt{x}(x+3))-\ln(x^2+1)=\ln(\frac{\sqrt{x}(x+3)}{x^2+1})$

16. satellite73

i made a mistake on the first try and wrote $\frac{1}{2}\ln(x)=\ln(x^2)$ which is wrong

17. satellite73

there now i deleted it. hope the steps are clear

18. animalsavior94

ohhh no i understand thanks so much!!:]

19. satellite73

welcome!