How do you solve $3{e^x} = 11 - {e^{ - x}}$ ?

Hint: In this question, we need to solve the given equation to find the value of x. Firstly, we will make rearrangement in the given equation to obtain a quadratic equation. Then we take ${e^x} = t$ and find the roots using the quadratic formula. After that we substitute ${e^x} = t$ and obtain the value for ${e^x}$. Since logarithm is an inverse function of exponential, we isolate ${e^x}$ by taking $\log $. We then simplify it and obtain the value for the variable x.

Complete step by step answer:
Given the equation of the form $3{e^x} = 11 - {e^{ - x}}$ …… (1)
We are asked to solve the above equation. i.e. to find the value for the variable x.
Firstly, we make rearrangements in the expression given in the equation (1) to simplify it
In the expression above, we rewrite ${e^{ - x}} = \dfrac{1}{{{e^x}}}$.
Hence equation (1) can be written as,
$ \Rightarrow 3{e^x} = 11 - \dfrac{1}{{{e^x}}}$
Now multiplying throughout by ${e^x}$ we get,
$ \Rightarrow 3{e^x} \cdot {e^x} = 11 \cdot {e^x} - \dfrac{1}{{{e^x}}} \cdot {e^x}$
$ \Rightarrow 3{e^{2x}} = 11{e^x} - 1$
Transferring all the terms to the L.H.S. we get,
$ \Rightarrow 3{e^{2x}} - 11{e^x} + 1 = 0$
Now we take $t = {e^x}$. So that it will be easier to solve. Hence we have,
$ \Rightarrow 3{t^2} - 11t + 1 = 0$ …… (2)
Note that the above equation is a quadratic equation.
If $a{x^2} + bx + c = 0$ is a quadratic equation, then we find the roots using quadratic formula given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Note that here $a = 3$, $b = - 11$ and $c = 1$.
Hence we get,
$t = \dfrac{{ - ( - 11) \pm \sqrt {{{( - 11)}^2} - 4 \times 3 \times 1} }}{{2 \times 3}}$
$ \Rightarrow t = \dfrac{{11 \pm \sqrt {121 - 12} }}{6}$
$ \Rightarrow t = \dfrac{{11 \pm \sqrt {109} }}{6}$
Now substituting back $t = {e^x}$ we get,
$ \Rightarrow {e^x} = \dfrac{{11 \pm \sqrt {109} }}{6}$
We know that logarithmic function is an inverse of exponential function. So to isolate ${e^x}$, we take $\ln $(natural logarithm) on both sides of the above equation we get,
$ \Rightarrow \ln {e^x} = \ln \left( {\dfrac{{11 \pm \sqrt {109} }}{6}} \right)$
Note that $\ln {e^x} = x$. Hence we have,
$ \Rightarrow x = \ln \left( {\dfrac{{11 \pm \sqrt {109} }}{6}} \right)$
Hence we get two values for the variable x which is given by,
$x = \ln \left( {\dfrac{{11 + \sqrt {109} }}{6}} \right)$ and $x = \ln \left( {\dfrac{{11 - \sqrt {109} }}{6}} \right)$

Hence, the solution for the equation $3{e^x} = 11 - {e^{- x}}$ is given by $x = \ln \left( {\dfrac{{11 + \sqrt {109} }}{6}} \right)$ and $x = \ln \left( {\dfrac{{11 - \sqrt {109} }}{6}} \right)$.

Note: Students must remember that logarithm is an inverse of exponential function. So to isolate exponential function and to obtain the value for the unknown variable we must take $\log $on both sides of the equation and solve it.
Also it is important to substitute ${e^x} $ as $t = {e^x}$, so that it will be easier to solve the problem.
Consider the general quadratic equation $a{x^2} + bx + c = 0$, then we find the roots using quadratic formula given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.