coolaidd
(1) Write log9+1/3log729 as a single logarithm.
(2) Write log[2]5-1/2log[2]169 as a single logarithm.
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jbovey
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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)
jbovey
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log(6561)^(1/3)
coolaidd
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so i solve that? @jbovey i got In9/in10
coolaidd
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is that for the #(2)?
jbovey
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no thats the answer for #1. Since it says just write as single log you wouldn't have to solve any further
jbovey
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it just wants you to use the log rules and combine
coolaidd
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oooh ok.. so what about #2?
jbovey
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Im kind of confused by the way you wrote it. Can u draw the equation out so I can see?
coolaidd
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yupp! hold on..
jbovey
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I know you'll have to use log rule 3 which is log(a)-log(b)= log(a)/(b)
coolaidd
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|dw:1354512532548:dw|
jbovey
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so are 5 and 169 the exponents?
coolaidd
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no those are regular sized..
geoffb
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The 2s are bases.
She means \(\log_{2} 5 - \frac{1}{2} \log_{2} 169\)
coolaidd
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25 and 338?
jbovey
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ohhhh sorry about that, thanks @geoffb
geoffb
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No problem. :)
jbovey
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wait so wouldnt that make 5 and 169 the exponents? @geoffb
geoffb
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No, it wouldn't.
geoffb
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You would just need to maintain base 2. It's nice because both logs use base 2.
campbell_st
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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...
geoffb
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So, like you said, you could start by moving the 1/2 up as an exponent.
$$\large \log_{2} 5 - \frac{1}{2} \log_{2} 169 = \log_{2} 5 - \log_{2} 169^{\frac{1}{2}} = \log_{2} (\frac{5}{13})$$
jbovey
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Thanks bro @campbell_st I was close lol
coolaidd
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ok..so would that be the final answer @geoffb ?
jbovey
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yeah thats the answer @coolaidd
coolaidd
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thanks!