Respuesta :
Given the equation:
[tex]e^x+e^{-x}=3[/tex]We have the following equivalent expression:
[tex]e^x+e^{-x}=e^x+\frac{1}{e^x}=\frac{(e^x)^2+1^{}}{e^x}^{}[/tex]then, substituting in the first equation, we have:
[tex]\frac{(e^x)^2+1}{e^x}=3[/tex]since e^x is always positive, we can multiply both sides by it to get:
[tex]\begin{gathered} (e^x)^2+1=3e^x \\ \Rightarrow(e^x)^2-3(e^x)+1=0^{} \end{gathered}[/tex]then, if we make u = e^x, we get the following quadratic equation:
[tex]\begin{gathered} u=e^x \\ \Rightarrow u^2-3u+1=0 \end{gathered}[/tex]which have the following solutions:
[tex]\begin{gathered} u_1=\frac{3_{}+\sqrt[]{5}}{2} \\ u_2=\frac{3-\sqrt[]{5}}{2} \end{gathered}[/tex]then, by reversing the substiution of u = e^x, we get:
[tex]\begin{gathered} e^x=\frac{3\pm\sqrt[]{5}}{2} \\ \end{gathered}[/tex]finally, using the natural logarithm on both sides of the equation, we get the following:
[tex]\begin{gathered} \ln (e^x)=\ln (\frac{3\pm\sqrt[]{5}}{2}) \\ \Rightarrow x=\ln (\frac{3\pm\sqrt[]{5}}{2}) \\ \Rightarrow x_1=\ln (\frac{3+\sqrt[]{5}}{2})=0.96 \\ \Rightarrow x_2=\ln (\frac{3-\sqrt[]{5}}{2})=-0.96 \end{gathered}[/tex]therefore, the solution of the exponential equation is x = ln(3 + sqrt(5)/2) = 0.96 and x = ln(3-sqrt(5)/2) = -0.96
