Question

Me ajude nessas questões

Answer 1

Resposta:

39: $\ln1215$,   40:   $\ln\left(\dfrac{a^2-b^2}{c^2}\right)$,   41:  $\ln\left(\dfrac{(x+2)\sqrt{2}}{x^2+3x+2}\right).$

Explicação passo-a-passo:

Olá!

     Lembre das propriedades de logaritmo:

$\bullet \displaystyle \ln (x\cdot y)=\ln x+\ln y\\\\ \bullet \ln\left(\dfrac{x}{y}\right)=\ln x - \ln y\\\\ \bullet \ln x^y=y\cdot \ln x.$

       Com elas em mente, temos:

$\text{\bf{39.} } \ln5+5\ln3=\ln5+\ln3^5=\ln5+\ln243=\ln(5\cdot 243)=\\=\ln1215.\\\\ \\\text{\bf{40.} } \ln(a+b)+\ln(a-b)-2\ln c=\ln[(a+b)\cdot(a-b)]-\ln c^2=\\=\ln(a^2-b^2)-\ln c^2=\ln\left(\dfrac{a^2-b^2}{c^2}\right).\\\\ \\\text{\bf{41.} } \dfrac{1}{3}\ln(x+2)^3+\dfrac{1}{2}\left[\ln x- \ln(x^2+3x+2)^2\right]=\\=\ln(x+2)^{\frac{1}{3}\cdot 3}+\ln x^{\frac{1}{2}}-\ln(x^2+3x+2)^{\frac{1}{2}\cdot 2}=\\=\ln(x+2) + \ln\sqrt{2}-\ln(x^2+3x+2)=\ln\left(\dfrac{(x+2)\sqrt{2}}{x^2+3x+2}\right).$

Bons estudos!