Question

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

Answer 1

We substitute the value of exponential as a variable and form an equation in the new variable. Use the concept of reciprocal and write the equation. Convert the equation into quadratic equation by taking LCM. Solve for the value of the variable using determinant formula. In the end substitute back the values assumed in the staring and take log on both sides to calculate the values of x.

• If the power has negative sign, then we take reciprocal of the term i.e. ${a^{ - 1}} = \dfrac{1}{a}$
• For a general quadratic equation $a{x^2} + bx + c = 0$, roots are given by formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step answer:
We are given the equation ${e^x} + {e^{ - x}} = 3$ … (1)
We have one term with negative power, so we can apply the concept of reciprocal and write ${e^{ - x}} = \dfrac{1}{{{e^x}}}$in equation (1)
$\Rightarrow {e^x} + \dfrac{1}{{{e^x}}} = 3$ … (2)
Let us substitute the value of ${e^x} = t$in equation (2)
$\Rightarrow t + \dfrac{1}{t} = 3$
Now we can take LCM on left hand side of the equation
$\Rightarrow \dfrac{{{t^2} + 1}}{t} = 3$
Cross multiply values from left hand side to right hand side of the equation
$\Rightarrow {t^2} + 1 = 3t$
Shift all values to left hand side of the equation
$\Rightarrow {t^2} - 3t + 1 = 0$
Since this is a quadratic equation in variable ‘t’; we can apply determinant formula to calculate value of ‘t’
We know for a general quadratic equation $a{x^2} + bx + c = 0$, roots are given by formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So, for quadratic equation${t^2} - 3t + 1 = 0$ roots will be given by,
$\Rightarrow t = \dfrac{{ - ( - 3) \pm \sqrt {{{( - 3)}^2} - 4 \times 1 \times 1} }}{{2 \times 1}}$
Solve the value under the root,
$\Rightarrow t = \dfrac{{3 \pm \sqrt {9 - 4} }}{2}$
$\Rightarrow t = \dfrac{{3 \pm \sqrt 5 }}{2}$
We get $t = \dfrac{{3 + \sqrt 5 }}{2}$ and $t = \dfrac{{3 - \sqrt 5 }}{2}$
Now we know that we substitute ${e^x} = t$, so we will substitute back this value in place of ‘t’.
We get ${e^x} = \dfrac{{3 + \sqrt 5 }}{2}$ and ${e^x} = \dfrac{{3 - \sqrt 5 }}{2}$
Now we have to calculate the value of x, so we take log on both sides of the equation
We get $\log \left( {{e^x}} \right) = \log \left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)$and $\log \left( {{e^x}} \right) = \log \left( {\dfrac{{3 - \sqrt 5 }}{2}} \right)$
Since log and exponential cancel each other,
We get $x = \log \left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)$and $x = \log \left( {\dfrac{{3 - \sqrt 5 }}{2}} \right)$

$\therefore$Solution of ${e^x} + {e^{ - x}} = 3$ is $x = \log \left( {\dfrac{{3 + \sqrt 5 }}{2}} \right)$and $x = \log \left( {\dfrac{{3 - \sqrt 5 }}{2}} \right)$

Note:
Many students make mistake of writing the solution just as the value of ${e^x}$ which is wrong as the equation ${e^x} + {e^{ - x}} = 3$ is in ‘x’ variable so the solution should be the values of ‘x’. Keep in mind logarithm function cancels the exponential function. Also, many students forget to change the sign of the values when shifting values from one side of the equation to another, always change the sign from positive to negative and vice-versa when shifting values.