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

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jboveyBest ResponseYou've already chosen the best response.0
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)
 one year ago

coolaiddBest ResponseYou've already chosen the best response.0
so i solve that? @jbovey i got In9/in10
 one year ago

coolaiddBest ResponseYou've already chosen the best response.0
is that for the #(2)?
 one year ago

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

jboveyBest ResponseYou've already chosen the best response.0
it just wants you to use the log rules and combine
 one year ago

coolaiddBest ResponseYou've already chosen the best response.0
oooh ok.. so what about #2?
 one year ago

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

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

coolaiddBest ResponseYou've already chosen the best response.0
dw:1354512532548:dw
 one year ago

jboveyBest ResponseYou've already chosen the best response.0
so are 5 and 169 the exponents?
 one year ago

coolaiddBest ResponseYou've already chosen the best response.0
no those are regular sized..
 one year ago

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

jboveyBest ResponseYou've already chosen the best response.0
ohhhh sorry about that, thanks @geoffb
 one year ago

jboveyBest ResponseYou've already chosen the best response.0
wait so wouldnt that make 5 and 169 the exponents? @geoffb
 one year ago

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

campbell_stBest ResponseYou've already chosen the best response.1
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...
 one year ago

geoffbBest ResponseYou've already chosen the best response.1
So, like you said, you could start by moving the 1/2 up as an exponent. $$\large \log_{2} 5  \frac{1}{2} \log_{2} 169 = \log_{2} 5  \log_{2} 169^{\frac{1}{2}} = \log_{2} (\frac{5}{13})$$
 one year ago

jboveyBest ResponseYou've already chosen the best response.0
Thanks bro @campbell_st I was close lol
 one year ago

coolaiddBest ResponseYou've already chosen the best response.0
ok..so would that be the final answer @geoffb ?
 one year ago

jboveyBest ResponseYou've already chosen the best response.0
yeah thats the answer @coolaidd
 one year ago
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