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