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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
 2 years ago
 2 years 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
 2 years ago
 2 years ago

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

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

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

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

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

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