## anonymous 5 years ago write as logarithm of a single quantity: 1/3[4log5(x+1)-3log5(3-x)+6log5x]

1. nowhereman

So how far do you get?

2. anonymous

not at all because its so confusing

3. anonymous

is this log base 5?

4. anonymous

actually it doesnt matter so lets just write log

5. anonymous

and ignore the 1/3 out front we will deal with it last

6. anonymous

yes its a base 5

7. anonymous

first use $\log(a^n)=n\log(a)$ backwards. all the multipliers come inside the log as exponents

8. anonymous

$4\log(x+1)=\log((x+1)^4)$ $3\log(3-x)=\log((3-x)^3)$ $6\log(x)=\log(x^6)$

9. anonymous

giving $\log((x+1)^4)-\log((3-x)^3)+\log(x^6)$

10. anonymous

now we use $\log(a)-log(b)=log(\frac{a}{b})$

11. anonymous

what happened to the 1/3 outside the []

12. anonymous

i will deal with that last

13. anonymous

not done yet!

14. anonymous

$log((x+1)^4)-\log((3-x)^3)=\log(\frac{(x+1)^4}{(3-x)^3})$

15. anonymous

now we use $\log(a)+\log(b)=\log(ab)$

16. anonymous

$\log(\frac{(x+1)^4}{(3-x)^3})+\log(x^6)=\log(\frac{x^6(x+1)^4}{(3-x)^3})$

17. anonymous

now for the 1/3 out front. take the cube root of all that stuff.

18. anonymous

the inner stuff i mean. the cube root of x^6 is x^2, and the cube root of (3-x)^3 is 3-x and so you get $\log(\frac{x^2(x+1)^{\frac{4}{3}}}{3-x})$

19. anonymous

wow that was a hard problem but thanks again:)