1. anonymous

$\log \left( \frac{ 10^{x} }{ x(x ^{2}+1)(x ^{4}+2) } \right)$

2. freckles

have you considered first using the quotient rule for log? $\log(\frac{u}{v})=\log(u)-\log(v) \\$

3. freckles

$\log(uv)=\log(u)+\log(v) \text{ is another rule you can use after that } \\ \log(u^r)=rlog(u) \text{ is the power rule which can also be used here }$

4. anonymous

$\log10^{x}-logx (x ^{2}+1)+\log(x ^{4}+2)$

5. freckles

I think you forgot to distribute the minus sign

6. anonymous

$xlog10-logx(x ^{2}+1)+\log(x ^{4}+2)$

7. anonymous

but since the denominator is looks like multiplication wouldn't be a addition?

8. freckles

$\log(\frac{10^x}{x(x^2+1)(x^4+2)}) \\ \log(10^x)-\log(x(x^2+1)(x^4+2)) \\ \log(10^x)-[\log(x(x^2+1)(x^4+2)) ] \\ \log(10^x)-[\log(x(x^2+1))+\log(x^4+2)]] \\ \log(10^x)-\log(x(x^2+1))-\log(x^4+2)$

9. freckles

though could also expand the log(x(x^2+1)) part too

10. anonymous

ooh so the signs changed because of the minus sign before the brackets?

11. freckles

yep that is called the distributive property

12. anonymous

oh ok.. sorry i suck on paying attention to small details.

13. anonymous

ok so could i expand more after log(x(x^2+1))−log(x^4+2)?

14. freckles

the log(x(x^2+1)) could be expanded more

15. freckles

notice x(x^2+1) is a product of x and (x^2+1)

16. anonymous

would i t be: Logx+log (x^2+1)

17. freckles

yes and don't forget we also have a - sign in front of the log(x(x^2+1) so you have: $\log(\frac{10^x}{x(x^2+1)(x^4+2)}) \\ \log(10^x)-\log(x(x^2+1)(x^4+2)) \\ \log(10^x)-[\log(x(x^2+1)(x^4+2)) ] \\ \log(10^x)-[\log(x(x^2+1))+\log(x^4+2)]] \\ \log(10^x)-\log(x(x^2+1))-\log(x^4+2) \\ \log(10^x)-[\log(x)+\log(x^2+1)]-\log(x^4+2)$

18. freckles

anyways yes log(10^x) =xlog(10) but assuming this is log base 10 you could also simplify the x log(10) more

19. anonymous

it will just become x right?

20. freckles

right

21. anonymous

since log 10=1

22. freckles

yes

23. anonymous

the result is: $x-\log (x)-\log (x ^{2}+1)-\log (x ^{4}-2)$

24. anonymous

?

25. freckles

well you change the sign between x^4 and 2 for some reason

26. anonymous

oh haha yeah.. oops

27. freckles

$\log(\frac{10^x}{x(x^2+1)(x^4+2)}) \\ \log(10^x)-\log(x(x^2+1)(x^4+2)) \\ \log(10^x)-[\log(x(x^2+1)(x^4+2)) ] \\ \log(10^x)-[\log(x(x^2+1))+\log(x^4+2)]] \\ \log(10^x)-\log(x(x^2+1))-\log(x^4+2) \\ \log(10^x)-[\log(x)+\log(x^2+1)]-\log(x^4+2) \\ x \log(10)-\log(x)-\log(x^2+1)-\log(x^4+2) \\ x(1)-\log(x)-\log(x^2+1)-\log(x^4+2) \\ x-\log(x)-\log(x^2+1)-\log(x^4+2)$ should be right if the log's are in base 10