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gfields444
Group Title
Which of the following is the simplified version of one–half log 3 x – 2 log 3 y + 4 log 3 z?
a)log 3 square root of x over y squared times z to the fourth power
b)log 3 square root of x times y squared over z to the fourth power
c)log 3 square root of x times z to the fourth power over y squared
d)log 3 x3yz
 one year ago
 one year ago
gfields444 Group Title
Which of the following is the simplified version of one–half log 3 x – 2 log 3 y + 4 log 3 z? a)log 3 square root of x over y squared times z to the fourth power b)log 3 square root of x times y squared over z to the fourth power c)log 3 square root of x times z to the fourth power over y squared d)log 3 x3yz
 one year ago
 one year ago

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Shi Group TitleBest ResponseYou've already chosen the best response.1
Are you familiar with the properties of logarithms? \[\log_{b} (m*n) = \log_{b} m + \log_{b}n \] \[\log_{b}(m/n) = \log_{b}m  \log_{b}n \] \[\log_{b}m ^{x} = x \log_{b} m \]
 one year ago

gfields444 Group TitleBest ResponseYou've already chosen the best response.0
yes but I tried each of the 4 potential answers and I keep getting the wrong answer. I think it might be the last one.
 one year ago

Shi Group TitleBest ResponseYou've already chosen the best response.1
Start by applying the third property to all applicable logs, so \[2\log_{3} y = \log_{3} y ^{2}\]
 one year ago

Shi Group TitleBest ResponseYou've already chosen the best response.1
ah let me check
 one year ago

Shi Group TitleBest ResponseYou've already chosen the best response.1
the answer is in the list, maybe your mistake was you got \[\log_{3}\frac{ \sqrt{x} }{ y ^{2}z ^{4} }\] ??
 one year ago

gfields444 Group TitleBest ResponseYou've already chosen the best response.0
I just tried it again.... the equations you gave me helped!! I see what I did wrong, of course it was simple miscalculation :p
 one year ago

Shi Group TitleBest ResponseYou've already chosen the best response.1
cool glad you got it :)
 one year ago

gfields444 Group TitleBest ResponseYou've already chosen the best response.0
thank you!
 one year ago
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