Chapter 4.5, Problem 31E

To determine

Solve the equation.

Answer to Problem 31E

  $\boxed{x\approx 0.5493}$

Explanation of Solution

Given information:

Solve the equation.

  $e^{4x}+4e^{2x}-21=0$

Calculation:

Consider the given equation,

  $e^{4x}+4e^{2x}-21=0$

Apply law of exponents,

  ${\left(e^{2x}\right)}^2+4e^{2x}-21=0$

Now this is a quadratic equation,

  $\begin{array}{l}{e}^{2x}=\frac{-b\pm \sqrt{b^2-4ac}}{2a},a=1,b=4,c=-21\\ e^{2x}=\frac{-\left(4\right)\pm \sqrt{{\left(4\right)}^2-4\times 1\times \left(-21\right)}}{2\times 1}\\ =\frac{-4\pm \sqrt{100}}{2}\\ =\frac{-4\pm 10}{2}=\frac{6}{2},\frac{-14}{2}\\ e^{2x}=3,-7\\ e^{2x}=3,\end{array}$

By taking log both sides,

  $\begin{array}{l}\ln e^{2x}=\ln 3,\\ 2x=\ln 3\\ x=\frac{\ln 3}{2}\\ x=0.5493061\end{array}$

  $\boxed{x\approx 0.5493}$

  $\boxed{e^{2x}=-7}$ has no solution.

Hence, the solution is, $\boxed{x\approx 0.5493}$