Question

How do you solve ${{e}^{x}}+{{e}^{-x}}=4$ ?

Answer 1

Hint: We are given an equation consisting of exponents raised to the power of functions of x-variable. In order to solve this equation and find the value of x, we must rearrange the terms by using the suitable mathematical operations. Then, we shall take ${{e}^{x}}$ as another variable to easily solve the obtained quadratic equation and later on substitute the values to get the final solution.

Complete step-by-step solution:
We are given the equation, ${{e}^{x}}+{{e}^{-x}}=4$. Here the negative power of exponent e denotes that this term should be in the denominator and the numerator must be 1 since the constant coefficient of this term is 1.
$\Rightarrow {{e}^{-x}}=\dfrac{1}{{{e}^{x}}}$
Modifying this term in the given equation, we get
$\Rightarrow {{e}^{x}}+\dfrac{1}{{{e}^{x}}}=4$
Taking LCM in the left-hand side, we get
\[\begin{align}
  & \Rightarrow \dfrac{{{e}^{x}}\left( {{e}^{x}} \right)+1}{{{e}^{x}}}=4 \\
 & \Rightarrow \dfrac{{{\left( {{e}^{x}} \right)}^{2}}+1}{{{e}^{x}}}=4 \\
 & \Rightarrow {{e}^{2x}}+1=4{{e}^{x}} \\
\end{align}\]
\[\Rightarrow {{e}^{2x}}-4{{e}^{x}}+1=0\]
Let ${{e}^{x}}=t$. Substituting this value, we get
$\Rightarrow {{t}^{2}}-4t+1=0$
To solve this quadratic equation, we shall first find its discriminant $\left( D \right)$ and then find the solutions of t.
For any quadratic equation, $ax+by+c=0$, the discriminant is given as:
$D={{b}^{2}}-4ac$
Here, we have $a=1,b=-4,c=1$
$\begin{align}
  & \Rightarrow D={{\left( -4 \right)}^{2}}-4\left( 1 \right)1 \\
 & \Rightarrow D=16-4 \\
 & \Rightarrow D=12 \\
\end{align}$
The solutions of the equation are expressed as:
$t=\dfrac{-b\pm \sqrt{D}}{2a}$
Substituting the values from our quadratic equation, we get
$\Rightarrow t=\dfrac{-\left( -4 \right)\pm \sqrt{12}}{2\left( 1 \right)}$
We know that 12 is not a perfect square and its square root is equal to $2\sqrt{3}$. Putting this value, we get
$\Rightarrow t=\dfrac{4\pm 2\sqrt{3}}{2}$
$\Rightarrow t=\dfrac{4+2\sqrt{3}}{2},\dfrac{4-2\sqrt{3}}{2}$
Therefore, ${{e}^{x}}=\dfrac{4+2\sqrt{3}}{2},\dfrac{4-2\sqrt{3}}{2}$.
This implies that ${{e}^{x}}=\dfrac{4+2\sqrt{3}}{2}$ and ${{e}^{x}}=\dfrac{4-2\sqrt{3}}{2}$.
Now, taking ${{\log }_{e}}$ on both these equations, we get
$\Rightarrow {{\log }_{e}}{{e}^{x}}={{\log }_{e}}\dfrac{4+2\sqrt{3}}{2}$ and $\Rightarrow {{\log }_{e}}{{e}^{x}}={{\log }_{e}}\dfrac{4-2\sqrt{3}}{2}$
According to a property of logarithms, we know that $\log {{a}^{b}}=b\log a$. Applying this property in our expression, we get
$\Rightarrow x{{\log }_{e}}e={{\log }_{e}}\dfrac{4+2\sqrt{3}}{2}$ and $\Rightarrow x{{\log }_{e}}e={{\log }_{e}}\dfrac{4-2\sqrt{3}}{2}$
Another logarithmic property says that if the base is equal to the argument of the logarithmic function, then it is equal to 1. It is given as ${{\log }_{a}}a=1$. This implies that ${{\log }_{e}}e=1$.
$\Rightarrow x.1={{\log }_{e}}\dfrac{4+2\sqrt{3}}{2}$ and $\Rightarrow x.1={{\log }_{e}}\dfrac{4-2\sqrt{3}}{2}$
$\therefore x={{\log }_{e}}\dfrac{4+2\sqrt{3}}{2}$ and $x={{\log }_{e}}\dfrac{4-2\sqrt{3}}{2}$
Therefore, the solution of ${{e}^{x}}+{{e}^{-x}}=4$ is $x={{\log }_{e}}\dfrac{4+2\sqrt{3}}{2},{{\log }_{e}}\dfrac{4-2\sqrt{3}}{2}$.

Note: While solving such problems, we must have prior knowledge of exponents, logarithmic functions and their properties. If we tend to apply any property in our solution in some wrong manner, then the entire answer can go wrong.