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zvonat [6]
2 years ago
8

Write as a single logarithm 3-1/2(log6+log3-3log2)

Mathematics
2 answers:
MA_775_DIABLO [31]2 years ago
8 0
Answer in document below.

faust18 [17]2 years ago
6 0
3-\dfrac{1}{2}\left(\log6+\log3-3\log2\right)=\\
\log10^3-\dfrac{1}{2}\left(\log\left(6\cdot3\right)-\log2^3\right)=\\
\log1000-\dfrac{1}{2}\left(\log18-\log8\right)=\\
\log1000-\dfrac{1}{2}\left(\log\dfrac{18}{8}\right)=\\
\log1000-\dfrac{1}{2}\left(\log\dfrac{9}{4}\right)=\\
\log1000-\log\left(\dfrac{9}{4}\right)^{\dfrac{1}{2}}=\\
\log1000-\log\sqrt{\dfrac{9}{4}}=\\
\log1000-\log\dfrac{3}{2}=\\
\log\dfrac{1000}{\frac{3}{2}}=\\
\log\left(1000\cdot\dfrac{2}{3}\right)=\\
\log\dfrac{2000}{3}
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Step-by-step explanation:

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3 0
1 year ago
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Vanyuwa [196]
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The the value of θ = 0.955 = 0.96 (to 2 d.p) radian

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