coolaidd 2 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. jbovey

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. jbovey

log(6561)^(1/3)

3. coolaidd

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

4. coolaidd

is that for the #(2)?

5. jbovey

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

6. jbovey

it just wants you to use the log rules and combine

7. coolaidd

oooh ok.. so what about #2?

8. jbovey

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

9. coolaidd

yupp! hold on..

10. jbovey

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

11. coolaidd

|dw:1354512532548:dw|

12. jbovey

so are 5 and 169 the exponents?

13. coolaidd

no those are regular sized..

14. coolaidd
15. geoffb

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

16. coolaidd

25 and 338?

17. jbovey

ohhhh sorry about that, thanks @geoffb

18. geoffb

No problem. :)

19. jbovey

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

20. geoffb

No, it wouldn't.

21. geoffb

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. geoffb

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. jbovey

Thanks bro @campbell_st I was close lol

25. coolaidd

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

26. jbovey