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Résoudre les équations suivantes dans $\mathbb R_+^*$.
a.
$\ln(x) = -3$;
Corrigé
\[\ln(x) = -3 \iff x = \mathrm e^{-3}.\]
Donc $S = \big\{\mathrm e^{-3}\big\}$.
b.
$2\ln(x)-1 = 0$;
Corrigé
\[\begin{aligned}
&2\ln(x) - 1 =0&
\\ \iff
&\ln(x) = \frac 1 2&
\\ \iff
&x = \mathrm e^{1/2}&
\\ \iff
&x = \sqrt{\mathrm e}&
\\
&S = \left\{\sqrt{\mathrm e}\right\}.&
\end{aligned}\]
c.
$\left(\ln(x) + 5\right)\left(4\ln(x)- 5\right) = 0$;
Corrigé
\[\begin{aligned}
&\left(\ln(x)+5\right)\left(4\ln(x)-5\right)=0&
\\ \iff
&\begin{cases}\ln(x) + 5 = 0\\ 4\ln(x)-5 = 0\end{cases}&
\\ \iff
&\begin{cases}\ln(x)=-5\\ \ln(x) = \frac 5 4\end{cases}&
\\ \iff
&\begin{cases}x=\mathrm e^{-5}\\ x=\mathrm e^{5/4}\end{cases}&
\\
&S=\big\{\mathrm e^{-5};\mathrm e^{5/4}\big\}.&
\end{aligned}\]
d.
$\left(\ln(x)\right)^2 = 9$.
Corrigé
\[\begin{aligned}
&\left(\ln(x)\right)^2 = 9&
\\ \iff
&\begin{cases}\ln(x) = 3\\ \ln(x) = -3\end{cases}&
\\ \iff
&\begin{cases}x = \mathrm e^3\\x = \mathrm e^{-3}\end{cases}&
\\
&S = \big\{\mathrm e^{-3};\mathrm e^3\big\}.&
\end{aligned}\]
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