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

Answer 1

Sadie Alexander

#color(blue)(ln(m)-1/2ln(m+3)#

#sqrt(m^2/(m+3))=(m^2/(m+3))^(1/2)#

#ln(m^2/(m+3))^(1/2)=1/2ln(m^2/(m+3))#

#ln(a/b)=ln(a)-ln(b)#

#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))#

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Answer 2

Xavier Alonso

To expand ( \ln(\sqrt{\frac{m^2}{m+3}}) ), first, use the property of logarithms: ( \ln(ab) = \ln(a) + \ln(b) ). Then, apply the rule of exponents: ( \sqrt{a} = a^{1/2} ).

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

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.