Corrigé du 65 P. 300

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a. \[\ln(x) = 6 \iff x = \mathrm e^6\] donc $S = \left\{\mathrm e^6\right\}$.

b. \[\ln(x) = -2 \iff x = \mathrm e^{-2}\] donc $S = \left\{\mathrm e^{-2}\right\}$.

c. \[\begin{aligned} 1 - 2\ln(x) &= 0& \\ \iff 1 &= 2\ln(x)& \\ \iff \dfrac 1 2 &= \ln(x)& \\ \iff \mathrm e^{1/2} &= x.& \end{aligned}\] Donc $S = \left\{\mathrm e^{1/2}\right\} = \left\{\sqrt{\mathrm e}\right\}$.

d. \[\begin{aligned} 3\ln(x)+11&= 5& \\ \iff 3\ln(x) &= -6& \\ \iff \ln(x) &= -2& \\ \iff &x = \mathrm e^{-2}.& \end{aligned}\] Donc $S = \left\{\mathrm e^{-2}\right\} = \left\{\dfrac 1 {\mathrm e^2}\right\}$.

e. \[\ln(x) - 0,1 = 0 \iff \ln(x) = 0,1 \iff x = \mathrm e^{0,1}.\] Donc $S = \left\{\mathrm e^{0,1}\right\}$.

f. \[\ln(4x) = 1 \iff 4x = \mathrm e^1 = \mathrm e \iff x = \dfrac{\mathrm e}4.\] Donc $S = \left\{\dfrac {\mathrm e}4\right\}$.

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