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Mathematics 17 Online
OpenStudy (darkbluechocobo):

Help with logarithms please

OpenStudy (darkbluechocobo):

OpenStudy (kropot72):

\[\large \ln \sqrt{\frac{m^{2}}{(m+3)}}=\ln [m(m+3)^{-\frac{1}{2}}]\] \[\large =\ln m-\frac{1}{2}\ln (m+3)\]

OpenStudy (anonymous):

\[\ln \sqrt{\frac{ m ^{2} }{ m+3 }} = \ln \left( \frac{ m ^{2} }{ m+3 } \right)^{\frac{ 1 }{ 2 }}\]

OpenStudy (darkbluechocobo):

Alright soh seeing how you placed your equation. Would \[\ln (\frac{ e^3 }{ xy}) \] be -3 - ln(x)+ln(y)?

OpenStudy (darkbluechocobo):

or should it be -ln(y)?

OpenStudy (anonymous):

\[= \frac{ 1 }{ 2 }\left( \ln \frac{ m ^{2} }{ m+3 }\right)\]

OpenStudy (anonymous):

would be the next step

OpenStudy (anonymous):

\[= \frac{ 1 }{ 2 }\left( \ln m ^{2}-\ln m+3 \right)\]

OpenStudy (anonymous):

\[\frac{ 1 }{ 2 }\left( 2\ln m-\ln m+3\right)\]

OpenStudy (darkbluechocobo):

1/2lnm^2 -1/2m + 1.5?

OpenStudy (anonymous):

using the distributive property \[\ln m-\frac{ 1 }{ 2 }\ln \left( m+3 \right)\]

OpenStudy (anonymous):

my final post would be the correct answer.

OpenStudy (darkbluechocobo):

Thank you

OpenStudy (anonymous):

that make sense?

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