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

Mathematics
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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)
log(6561)^(1/3)
so i solve that? @jbovey i got In9/in10

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Other answers:

is that for the #(2)?
no thats the answer for #1. Since it says just write as single log you wouldn't have to solve any further
it just wants you to use the log rules and combine
oooh ok.. so what about #2?
Im kind of confused by the way you wrote it. Can u draw the equation out so I can see?
yupp! hold on..
I know you'll have to use log rule 3 which is log(a)-log(b)= log(a)/(b)
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so are 5 and 169 the exponents?
no those are regular sized..
http://www.mathway.com/math_image.aspx?p=logSMB02SSMB032SMB02sSMB03(5)-SMB02FSMB031SMB102SMB02fSMB03logSMB02SSMB032SMB02sSMB03(169)?p=163?p=42
The 2s are bases. She means \(\log_{2} 5 - \frac{1}{2} \log_{2} 169\)
25 and 338?
ohhhh sorry about that, thanks @geoffb
No problem. :)
wait so wouldnt that make 5 and 169 the exponents? @geoffb
No, it wouldn't.
You would just need to maintain base 2. It's nice because both logs use base 2.
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...
So, like you said, you could start by moving the 1/2 up as an exponent. $$\large \log_{2} 5 - \frac{1}{2} \log_{2} 169 = \log_{2} 5 - \log_{2} 169^{\frac{1}{2}} = \log_{2} (\frac{5}{13})$$
Thanks bro @campbell_st I was close lol
ok..so would that be the final answer @geoffb ?
yeah thats the answer @coolaidd
thanks!

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