Desenvolva.
\ln \left(\frac{a^{2}}{b \sqrt{c}}\right)

Passo 1

O enunciado pede expandir o logaritmo
\ln\!\bigl(\tfrac{a^{2}}{b\sqrt{c}}\bigr).
Vamos usar as propriedades de logaritmos: \ln(x/y)=\ln(x)-\ln(y), \ln(xy)=\ln(x)+\ln(y) e \ln(x^{k})=k\ln(x), além de \sqrt{c}=c^{1/2}.

Passo 2

Aplicando \ln(x/y)=\ln(x)-\ln(y) a \ln\!\bigl(\tfrac{a^{2}}{b\sqrt{c}}\bigr), temos
\ln\!\bigl(\tfrac{a^{2}}{b\sqrt{c}}\bigr) =\ln\bigl(a^{2}\bigr)-\ln\bigl(b\sqrt{c}\bigr).

Passo 3

Em seguida, aplicamos \ln(xy)=\ln(x)+\ln(y) a \ln\bigl(b\sqrt{c}\bigr):
\ln\bigl(b\sqrt{c}\bigr) =\ln(b)+\ln\bigl(\sqrt{c}\bigr).

Passo 4

Finalmente, usamos \ln(a^{2})=2\ln(a) e \ln(\sqrt{c})=\ln(c^{1/2})=\tfrac12\ln\text{(c)}. Assim,
\ln\!\bigl(\tfrac{a^{2}}{b\sqrt{c}}\bigr) =2\ln(a)-\bigl[\ln(b)+\tfrac12\ln\text{(c)}\bigr] =2\ln(a)-\ln(b)-\tfrac12\ln\text{(c)}.

Resposta

2\ln(a)-\ln(b)-\tfrac12\ln\text{(c)}

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