How do you expand $ln \sqrt{\frac{m^{2}}{m + 3}}$ $ln sqrt(m^2/(m+3))$ ?

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Answer 1 · 2018-04-06T14:15:14

$ln (m) - \frac{1}{2} ln (m + 3)$ $color(blue)(ln(m)-1/2ln(m+3)$

$\sqrt{\frac{m^{2}}{m + 3}} = {(\frac{m^{2}}{m + 3})}^{\frac{1}{2}}$ $sqrt(m^2/(m+3))=(m^2/(m+3))^(1/2)$

${ln (\frac{m^{2}}{m + 3})}^{\frac{1}{2}} = \frac{1}{2} ln (\frac{m^{2}}{m + 3})$ $ln(m^2/(m+3))^(1/2)=1/2ln(m^2/(m+3))$

$ln (\frac{a}{b}) = ln (a) - ln (b)$ $ln(a/b)=ln(a)-ln(b)$

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

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1/2[2ln(m)-ln(m+3)]$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =color(blue)(ln(m)-1/2ln(m+3))$

How do you expand ln sqrt(m^2/(m+3))ln√m2m+3ln sqrt(m^2/(m+3))?