Corrigé du 68 P. 300

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a. $$\begin{array}{rlcr}& \ln (21)=\ln (3\times 7)=\ln (3)+\ln (7).& & \end{array}$$

b. $$\begin{array}{rlcr}& \ln (63)=\ln (9\times 7)=\ln (3^2\times 7)=2\ln (3)+\ln (7).& & \end{array}$$

c. $$\begin{array}{rlcr}& \ln \left(3\sqrt{21}\right)+\ln (49)=\ln (3)+\ln \left(\sqrt{21}\right)+\ln (7^2)& & \\ & =\ln (3)+\frac{1}{2}\ln (21)+2\ln (7)& & \\ & =\ln (3)+\frac{1}{2}\ln (3\times 7)+2\ln (7)& & \\ & =\ln (3)+\frac{1}{2}(\ln (3)+\ln (7))+2\ln (7)& & \\ & =\ln (3)+\frac{1}{2}\ln (3)+\frac{1}{2}\ln (7)+2\ln (7)& & \\ & =\frac{3}{2}\ln (3)+\frac{5}{2}\ln (7).& & \end{array}$$

d. $$\begin{array}{rlcr}& \ln \left(\frac{3}{7}\right)=\ln (3)-\ln (7).& & \end{array}$$

e. $$\begin{array}{rlcr}& \ln \left(\frac{343}{27}\right)=\ln (343)-\ln (27)=\ln (7^3)-\ln (3^3)& & \\ & =3\ln (7)-3\ln (3).& & \end{array}$$

f. $$\begin{array}{rlcr}& 4\ln \left(\frac{7}{3}\right)+\ln (7^2)=4(\ln (7)-\ln (3))+2\ln (7)& & \\ & =4\ln (7)-4\ln (3)+2\ln (7)=6\ln (7)-4\ln (3).& & \end{array}$$

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