How do you solve ${{e}^{x}}+{{e}^{-x}}=3$ ?

Hint: Firstly, we begin to solve the problem by assuming ${{e}^{x}}$ as $y$ and also the value of ${{e}^{-x}}$ then becomes $\dfrac{1}{y}$ . The expression obtained in terms of variable y is simplified further to obtain the quadratic equation. The roots of the equation can be found out by using the roots for a quadratic equation formula. We get two values from the formula both of which are the roots of the equation. Now we substitute these roots back into $y$ and then solve for $x$ .

Complete step by step solution:
The given expression is ${{e}^{x}}+{{e}^{-x}}=3$
Let us denote the exponential function ${{e}^{x}}$ in the above expression with the variable y which means,
$\Rightarrow y={{e}^{x}}$
The law of exponents states that any term with a negative exponent can be written as,
$\Rightarrow {{a}^{-x}}=\dfrac{1}{{{a}^{x}}}$
Obeying the same,
${{e}^{-x}}$ can be written as $\dfrac{1}{{{e}^{x}}}$
Since we have substituted ${{e}^{x}}=y$ then upon Substituting the same we get,
$\Rightarrow {{e}^{-x}}=\dfrac{1}{y}$
Now we have to rewrite the same expression in terms of $y$
The expression obtained will then be given as,
$\Rightarrow y+\dfrac{1}{y}=3$
Now let us Multiply $y$ on both sides of the expression.
$\Rightarrow {{y}^{2}}+\dfrac{y}{y}=3y$
$\Rightarrow {{y}^{2}}+1=3y$
Now we will be shifting all the terms towards the left-hand side to obtain a quadratic equation in $y$
$\Rightarrow {{y}^{2}}-3y+1=0$
The roots of a quadratic equation( $a{{x}^{2}}+bx+c=0$ ) can be found by the formula,
$x=\left[ \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \right]$
Finding out the roots of the equation using the quadratic formula stated above, we get,
$\Rightarrow y=\left[ \dfrac{-\left( -3 \right)\pm \sqrt{{{3}^{2}}-4\times 1\times 1}}{2\times 1} \right]$
Now let us first solve the expression inside the root.
$\Rightarrow y=\left[ \dfrac{3\pm \sqrt{9-4}}{2\times 1} \right]$
$\Rightarrow y=\left[ \dfrac{3\pm \sqrt{5}}{2\times 1} \right]$
Now let us further simplify the expression.
$\Rightarrow y=\left[ \dfrac{3\pm \sqrt{5}}{2} \right]$
Hence the roots of the expression in $y$ will be $y=\left[ \dfrac{3+\sqrt{5}}{2} \right],\left[ \dfrac{3-\sqrt{5}}{2} \right]$
We know that $y={{e}^{x}}$ , upon replacing the value of $y$ we obtain,
$\Rightarrow {{e}^{x}}=\left[ \dfrac{3\pm \sqrt{5}}{2} \right]$
According to the properties of logarithms, if ${{e}^{x}}=n$ then $x=\log n$ or $\ln n$
Following the same, the expression can be written as,
$\Rightarrow x=\ln \left[ \dfrac{3\pm \sqrt{5}}{2} \right]$
Hence the values of $x$ will then be,
$\Rightarrow x=\ln \left[ \dfrac{3+\sqrt{5}}{2} \right],\ln \left[ \dfrac{3-\sqrt{5}}{2} \right]$
It is better to leave these values as it is rather than evaluating because it is easier to solve the expression when we put these back into the question.

Note: The solution can be cross-checked by simply placing back the values of $x$ back into the expression which is given in the question, ${{e}^{x}}+{{e}^{-x}}=3$ .While cross-checking when you place the value back in $x$ , the $\ln$ and $e$ then get cancelled. So, it will be easier if we leave the values of x as $\ln \left[ \dfrac{3+\sqrt{5}}{2} \right],\ln \left[ \dfrac{3-\sqrt{5}}{2} \right]$ rather than evaluating it.