anonymous
  • anonymous
(1) Write log9+1/3log729 as a single logarithm. (2) Write log[2]5-1/2log[2]169 as a single logarithm.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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)
anonymous
  • anonymous
log(6561)^(1/3)
anonymous
  • anonymous
so i solve that? @jbovey i got In9/in10

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anonymous
  • anonymous
is that for the #(2)?
anonymous
  • anonymous
no thats the answer for #1. Since it says just write as single log you wouldn't have to solve any further
anonymous
  • anonymous
it just wants you to use the log rules and combine
anonymous
  • anonymous
oooh ok.. so what about #2?
anonymous
  • anonymous
Im kind of confused by the way you wrote it. Can u draw the equation out so I can see?
anonymous
  • anonymous
yupp! hold on..
anonymous
  • anonymous
I know you'll have to use log rule 3 which is log(a)-log(b)= log(a)/(b)
anonymous
  • anonymous
|dw:1354512532548:dw|
anonymous
  • anonymous
so are 5 and 169 the exponents?
anonymous
  • anonymous
no those are regular sized..
anonymous
  • anonymous
http://www.mathway.com/math_image.aspx?p=logSMB02SSMB032SMB02sSMB03(5)-SMB02FSMB031SMB102SMB02fSMB03logSMB02SSMB032SMB02sSMB03(169)?p=163?p=42
anonymous
  • anonymous
The 2s are bases. She means \(\log_{2} 5 - \frac{1}{2} \log_{2} 169\)
anonymous
  • anonymous
25 and 338?
anonymous
  • anonymous
ohhhh sorry about that, thanks @geoffb
anonymous
  • anonymous
No problem. :)
anonymous
  • anonymous
wait so wouldnt that make 5 and 169 the exponents? @geoffb
anonymous
  • anonymous
No, it wouldn't.
anonymous
  • anonymous
You would just need to maintain base 2. It's nice because both logs use base 2.
campbell_st
  • 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...
anonymous
  • anonymous
So, like you said, you could start by moving the 1/2 up as an exponent. $$\large \log_{2} 5 - \frac{1}{2} \log_{2} 169 = \log_{2} 5 - \log_{2} 169^{\frac{1}{2}} = \log_{2} (\frac{5}{13})$$
anonymous
  • anonymous
Thanks bro @campbell_st I was close lol
anonymous
  • anonymous
ok..so would that be the final answer @geoffb ?
anonymous
  • anonymous
yeah thats the answer @coolaidd
anonymous
  • anonymous
thanks!

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