## IloveCharlie Group Title Please help! I'm having trouble with logs. Click attached :) one year ago one year ago

1. IloveCharlie Group Title

2. sasogeek Group Title

do you know the laws of logs?

3. IloveCharlie Group Title

I'm faimiliar with them, yes

4. IloveCharlie Group Title

*familiar

5. sasogeek Group Title

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 Group Title

are we okay?

7. IloveCharlie Group Title

Yes, I understand. Thanks!

8. sasogeek Group Title

you're welcome :)

9. sasogeek Group Title

what did you get though?

10. Hero Group Title

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

11. Hero Group Title

$$\text{Latex is working}$$

12. sasogeek Group Title

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

13. sasogeek Group Title

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

14. IloveCharlie Group Title

:/ I'm having trouble...

15. sasogeek Group Title

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

16. IloveCharlie Group Title

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

17. IloveCharlie Group Title

@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. bamkey360 Group Title

u said what it equal to

19. sasogeek Group Title

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 Group Title

are we good to go?

21. IloveCharlie Group Title

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

22. sasogeek Group Title

no problem :) you're welcome x