## anonymous 3 years ago (1) Write log9+1/3log729 as a single logarithm. (2) Write log[2]5-1/2log[2]169 as a single logarithm.

1. anonymous

1) the 1/3 would just be the exponent of log(729) so log(729)^1/3 and then we use log rule log(ab)= log(a)+log(b) to get log(9)(729)^(1/3)

2. anonymous

log(6561)^(1/3)

3. anonymous

so i solve that? @jbovey i got In9/in10

4. anonymous

is that for the #(2)?

5. anonymous

no thats the answer for #1. Since it says just write as single log you wouldn't have to solve any further

6. anonymous

it just wants you to use the log rules and combine

7. anonymous

oooh ok.. so what about #2?

8. anonymous

Im kind of confused by the way you wrote it. Can u draw the equation out so I can see?

9. anonymous

yupp! hold on..

10. anonymous

I know you'll have to use log rule 3 which is log(a)-log(b)= log(a)/(b)

11. anonymous

|dw:1354512532548:dw|

12. anonymous

so are 5 and 169 the exponents?

13. anonymous

no those are regular sized..

14. anonymous
15. anonymous

The 2s are bases. She means $$\log_{2} 5 - \frac{1}{2} \log_{2} 169$$

16. anonymous

25 and 338?

17. anonymous

ohhhh sorry about that, thanks @geoffb

18. anonymous

No problem. :)

19. anonymous

wait so wouldnt that make 5 and 169 the exponents? @geoffb

20. anonymous

No, it wouldn't.

21. anonymous

You would just need to maintain base 2. It's nice because both logs use base 2.

22. campbell_st

just a quick point on your first question... $\frac{1}{3}\log(729) = \log(\sqrt[3]{729}) = \log(9)$ so then $\log(9) + \frac{1}{3}\log(729) = \log(9) + \log(9) = \log(9 \times 9) = \log(81).. or... 2\log(9)$ just a suggestion... when compared to $\log(\sqrt[3]{6561})$ not quite the same things...

23. anonymous

So, like you said, you could start by moving the 1/2 up as an exponent. $$\large \log_{2} 5 - \frac{1}{2} \log_{2} 169 = \log_{2} 5 - \log_{2} 169^{\frac{1}{2}} = \log_{2} (\frac{5}{13})$$

24. anonymous

Thanks bro @campbell_st I was close lol

25. anonymous

ok..so would that be the final answer @geoffb ?

26. anonymous