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let log(D^2/C^3*A) when logA=3 logB=2 logD=5 a)0.549 b)4.44 c)1 d)19
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\[\begin{align*}\log\left(\frac{D^2}{C^3A}\right)&=\log D^2-\log\left(C^3A\right)\\ &=2\log D-\left(\log C^3+\log A\right)\\ &=2\log D-3\log C-\log A\\ &=\cdots \end{align*}\]
i did that but i got none of the four answers that they give you. please explain more
Do you know what \(\log C\) is?
logC is 2
Ah, so that was a typo.. \[\begin{align*}\log\left(\frac{D^2}{C^3A}\right)&=\log D^2-\log\left(C^3A\right)\\ &=2\log D-\left(\log C^3+\log A\right)\\ &=2\log D-3\log C-\log A\\ &=2(5)-3(2)-3\\ &=10-6-3\\ &=\cdots \end{align*}\]
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yeah it was. thanks for the help
you're welcome
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