## anonymous 5 years ago 2log6+log (1/3)=log12 which properties of logarithmics are/is used? power property and product property quotient property and product property quotient property only power property only

1. anonymous

Well If I asked you what 2log6 + log(1/3) was, how would you solve it (without using a calculator)?

2. anonymous

my teacher didnt go over this but still put it on the end of course exam

3. anonymous

log (1/3) is saying -log 3. With that we can use quotient property

4. anonymous

If it's just 2log6 +log(1/3) you can reduce it down to 2log6-log(3) because log(1) equal zero

5. anonymous

2log6 can be written as log6^2 so, log 36

6. anonymous

log36/log 3 is log 12

7. anonymous

No it's not. log36 - log3 is log12.

8. anonymous

same thing

9. anonymous

make sense?

10. anonymous

no it isn't. $\frac{log\ 36}{log\ 3} \ne log(\frac{36}{3})$

11. anonymous

log36-log3 =log36/log3

12. anonymous

i'm not saying log(36/3)

13. anonymous

No, it doesn't. log 36 - log 3 = log(36/3)

14. anonymous

>>> log(36, 10) - log(3, 10) 1.0791812460476247 >>> log(36,10)/log(3,10) 3.2618595071429146

15. anonymous

Very different numbers.

16. anonymous

try it in your calculator, (or wolframalpha)

17. anonymous

log(36/3) is log(12) i checked with calculator and got the same answer so sorry for confusion just disregard the log36/log3

18. anonymous

You can see quite trivially that it will be equal because 36/3 is 12. So log of 36/3 would be the log of 12. The problem is more about finding that log(36) - log(3) can be written as log(36/3)

19. anonymous

ok so what was the problem?

20. anonymous

The problem with your answer was that you said log(36) - log(3) = log(36)/log(3). But the simplest solution is to just say: $2log6+log (1/3)$ $=log36 + log(\frac{1}{3})$ $=log (36 \times \frac{1}{3})$ $=log(12)$ Which just uses the power and product rules. That said though you could also solve it by just using the product rule, or using the quotient rule and the power rule. So there's no non-ambiguous way to answer the multiple choice.