## heathernelly Group Title ;))) one year ago one year ago

1. jim_thompson5910 Group Title

Similar Problem: $\Large \log_{3}(60) - \log_{3}(12)$ $\Large \log_{3}\left(\frac{60}{12}\right)$ $\Large \log_{3}(5)$ So $\Large \log_{3}(60) - \log_{3}(12) = \log_{3}(5)$

2. heathernelly Group Title

that's all i have to do?

3. jim_thompson5910 Group Title

pretty much, but just use your numbers in place of mine

4. heathernelly Group Title

yeah, thanks! do you know how to do the other one?

5. jim_thompson5910 Group Title

in general $\Large \log_{b}(x) - \log_{b}(y) = \log_{b}\left(\frac{x}{y}\right)$

6. jim_thompson5910 Group Title

what do 16 and 64 have in common

7. heathernelly Group Title

they both have a 6 in them, both even #'s

8. jim_thompson5910 Group Title

good, but in terms of factors

9. heathernelly Group Title

2^6? i don't know :S :/

10. jim_thompson5910 Group Title

both are powers of 2 or powers of 4

11. heathernelly Group Title

ahh, yeah

12. jim_thompson5910 Group Title

so 16 = 2^4 64 = 2^6 and 16 = 4^1 64 = 4^3

13. heathernelly Group Title

that's it?

14. jim_thompson5910 Group Title

well you need to rewrite both sides to have the same base

15. jim_thompson5910 Group Title

for instance, say you had 2^x = 16 you would rewrite 16 as 2^4 to get 2^x = 2^4 and you can see that x must be 4 (since both bases are 2)

16. jim_thompson5910 Group Title

so you're using this idea to solve the second problem

17. heathernelly Group Title

ohhh, well that's pretty simple! thank youu

18. heathernelly Group Title

actually waait, it's not! haha.. i just got confused

19. jim_thompson5910 Group Title

with what

20. jim_thompson5910 Group Title

where are you stuck

21. heathernelly Group Title

and you can see that x must be 4 (since both bases are 2) ??

22. jim_thompson5910 Group Title

well if b^x = b^y, then x = y

23. heathernelly Group Title

i'm confused because you did 2 different problems..:/

24. jim_thompson5910 Group Title

that's just an example

25. heathernelly Group Title

okay, that makes sense .. but i'm still confused on what to do after what you said ''16 = 2^4 64 = 2^6 and 16 = 4^1 64 = 4^3''

26. jim_thompson5910 Group Title

oh i meant to say 16 = 4^2, my bad

27. jim_thompson5910 Group Title

1/16=64^(4x-3) 1/(4^2) = (4^3)^(4x-3) ... since 16 = 4^2 and 64 = 4^3 4^(-2) = 4^(3*(4x-3)) so because the bases are now the same, we can say that the exponents must be the same so... -2 = 3*(4x-3) solve for x

28. heathernelly Group Title

ahh, got it :) thanks a lot!!!

29. jim_thompson5910 Group Title

np