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coolaidd
 2 years ago
(1) Write log9+1/3log729 as a single logarithm.
(2) Write log[2]51/2log[2]169 as a single logarithm.
coolaidd
 2 years ago
(1) Write log9+1/3log729 as a single logarithm. (2) Write log[2]51/2log[2]169 as a single logarithm.

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jbovey
 2 years ago
Best ResponseYou've already chosen the best response.01) 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)

coolaidd
 2 years ago
Best ResponseYou've already chosen the best response.0so i solve that? @jbovey i got In9/in10

jbovey
 2 years ago
Best ResponseYou've already chosen the best response.0no thats the answer for #1. Since it says just write as single log you wouldn't have to solve any further

jbovey
 2 years ago
Best ResponseYou've already chosen the best response.0it just wants you to use the log rules and combine

coolaidd
 2 years ago
Best ResponseYou've already chosen the best response.0oooh ok.. so what about #2?

jbovey
 2 years ago
Best ResponseYou've already chosen the best response.0Im kind of confused by the way you wrote it. Can u draw the equation out so I can see?

jbovey
 2 years ago
Best ResponseYou've already chosen the best response.0I know you'll have to use log rule 3 which is log(a)log(b)= log(a)/(b)

jbovey
 2 years ago
Best ResponseYou've already chosen the best response.0so are 5 and 169 the exponents?

coolaidd
 2 years ago
Best ResponseYou've already chosen the best response.0no those are regular sized..

geoffb
 2 years ago
Best ResponseYou've already chosen the best response.1The 2s are bases. She means \(\log_{2} 5  \frac{1}{2} \log_{2} 169\)

jbovey
 2 years ago
Best ResponseYou've already chosen the best response.0ohhhh sorry about that, thanks @geoffb

jbovey
 2 years ago
Best ResponseYou've already chosen the best response.0wait so wouldnt that make 5 and 169 the exponents? @geoffb

geoffb
 2 years ago
Best ResponseYou've already chosen the best response.1You would just need to maintain base 2. It's nice because both logs use base 2.

campbell_st
 2 years ago
Best ResponseYou've already chosen the best response.1just 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
 2 years ago
Best ResponseYou've already chosen the best response.1So, 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
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks bro @campbell_st I was close lol

coolaidd
 2 years ago
Best ResponseYou've already chosen the best response.0ok..so would that be the final answer @geoffb ?

jbovey
 2 years ago
Best ResponseYou've already chosen the best response.0yeah thats the answer @coolaidd
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