## animalsavior94 Group Title write as logarithm of a single quantity: 1/3[4log5(x+1)-3log5(3-x)+6log5x] 3 years ago 3 years ago

1. nowhereman

So how far do you get?

2. animalsavior94

not at all because its so confusing

3. satellite73

is this log base 5?

4. satellite73

actually it doesnt matter so lets just write log

5. satellite73

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

6. animalsavior94

yes its a base 5

7. satellite73

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

8. satellite73

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

9. satellite73

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

10. satellite73

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

11. animalsavior94

what happened to the 1/3 outside the []

12. satellite73

i will deal with that last

13. satellite73

not done yet!

14. satellite73

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

15. satellite73

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

16. satellite73

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

17. satellite73

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

18. satellite73

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. animalsavior94

wow that was a hard problem but thanks again:)