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ln3x-3lnx+ln3y
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I'm not too sure about what you exactly need. Are you meant to just simplify?
\[\ln 3x -3 \ln x + \ln 3y =\ln 3 + \ln x - 2\ln x + \ln 3 + \ln y\]
\[= \log \frac{9y}{x^2}\]
\[=2\ln 3 -2\ln x +\ln y\]
Mistake in the first expression I typed: -2lnx should be -3ln x as in the question.
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\[=\ln 3^2-\ln x^2+\ln y= \ln \frac{3^2}{x^2}+\ln y = \ln \frac{3^2y}{x^2}=\ln \frac{9y}{x^2}\]
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