Question

Solve the following exponential equation e^x + e^-x = 3

Answer 1

Given the equation:

`e^x+e^(-x)=3`

We have the following equivalent expression:

`e^x+e^(-x)=e^x+(1)/(e^x)=\frac{(e^x)^2+1^{}}{e^x}^{}`

then, substituting in the first equation, we have:

`((e^x)^2+1)/(e^x)=3`

since e^x is always positive, we can multiply both sides by it to get:

`\begin{gathered} (e^x)^2+1=3e^x \\ \Rightarrow(e^x)^2-3(e^x)+1=0^{} \end{gathered}`

then, if we make u = e^x, we get the following quadratic equation:

`\begin{gathered} u=e^x \\ \Rightarrow u^2-3u+1=0 \end{gathered}`

which have the following solutions:

`\begin{gathered} u_1=\frac{3_{}+\sqrt[]{5}}{2} \\ u_2=\frac{3-\sqrt[]{5}}{2} \end{gathered}`

then, by reversing the substiution of u = e^x, we get:

`\begin{gathered} e^x=\frac{3\pm\sqrt[]{5}}{2} \\ \end{gathered}`

finally, using the natural logarithm on both sides of the equation, we get the following:

`\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}`

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