Question

solve the equation e^x+e^-x=3

Answer 1

That is equivalent to the following:\[e^x+\dfrac{1}{e^x}=3\]We can then simplify as follows:\[\dfrac{e^{2x}+1}{e^x}=3\]\[e^{2x}+1=3e^x\]\[e^{2x}-3e^x+1=0\]\[\left(2e^x-1\right)\left(e^x-1\right)=0\]\[e^x=\left\{1, \dfrac{1}{2}\right\}\]\[x=\left\{\ln(1), \ln\left(\dfrac{1}{2}\right)\right\}\]\[x=\left\{0, -\ln 2\right\}\]

Answer 2

\[e^{2x}-3e^{x}+1=0\]
\[e^x=\frac{3\pm\sqrt{9-4}}{2}=\frac{3\pm\sqrt{5}}{2}\quad\Rightarrow\quad x_{1,2}=\ln{\frac{3\pm\sqrt{5}}{2}}\]