1. anonymous

2. sasogeek

do you know the laws of logs?

3. anonymous

I'm faimiliar with them, yes

4. anonymous

*familiar

5. sasogeek

ok so one of them says that when you have log(a/b), it is the same as log(a)-log(b) ... you can apply that hear :)

6. sasogeek

are we okay?

7. anonymous

Yes, I understand. Thanks!

8. sasogeek

you're welcome :)

9. sasogeek

what did you get though?

10. Hero

Didn't you post this same problem the other day @IloveCharlie

11. Hero

$$\text{Latex is working}$$

12. sasogeek

/(/large hi /) ... if this doesn't appear correctly, then i probably forgot how to write latex, if not, yay! else, :(

13. sasogeek

ok so it messed up.... what did i do wrong?

14. anonymous

:/ I'm having trouble...

15. sasogeek

$$found \ what \ i \ did \ wrong \ \text{:P}$$ anyway what are you having trouble with?

16. anonymous

Is it log(3.8)-log(5)?

17. anonymous

@Hero Hmm... no. If so, that'd be weird. I don't think my Precal/Trig online course would repeat problems. Maybe it was something similar?

18. anonymous

u said what it equal to

19. sasogeek

actually, no. the question has a quotient/division thingy, which obviously means there's two sides to it. one is $$\large \sqrt{(3.2)(4.7)}$$ and the other is $$\large 5$$ according to the law, $$\large log(\frac{a}{b})=log(a) - log(b)$$ in this case, a= $$\large \sqrt{(3.2)(4.7)}$$ and b = $$\large 5$$ hence: $$\large log(\frac{\sqrt{(3.2)(4.7)}}{5})=log(\sqrt{(3.2)(4.7)}) - log(5)$$ =$$\large log(\sqrt{(15.04)} \ )-log(5)$$

20. sasogeek

are we good to go?

21. anonymous

Ohhhh o.k. Now I get it. Thanks so much sasogeek

22. sasogeek

no problem :) you're welcome x