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Sagot :
In order to rewrite this equation into a quadratic equation, let's substitute the term e^x by a new variable: y. So we have:
[tex]\begin{gathered} 2e^{2x}+e^x=15 \\ 2(e^x)^2+e^x=15 \\ 2y^2+y=15 \\ 2y^2+y-15=0 \end{gathered}[/tex]Now, solving this equation using the quadratic formula, we have:
[tex]\begin{gathered} a=2,b=1,c=-15 \\ y_1=\frac{-b+\sqrt[]{b^2-4ac}}{2a}=\frac{-1+\sqrt[]{1+120}}{4}=\frac{-1+11}{4}=\frac{5}{2} \\ y_2=\frac{-b-\sqrt[]{b^2-4ac}}{2a}-\frac{-1-11}{4}=-3 \end{gathered}[/tex]Now, going back to the substitution y = e^x, we have:
[tex]\begin{gathered} e^x=\frac{5}{2} \\ \ln (e^x)=\ln (\frac{5}{2}) \\ x=0.9163 \\ \\ e^x=-3 \\ x\text{ does not exist} \end{gathered}[/tex]So the solution to the initial equation is x = ln(5/2)
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