Chapter 0.5, Problem 36E

To determine

To solve: the given equation.

Answer to Problem 36E

The solutions are $x=-2.26,2.26$ .

Explanation of Solution

Given:

  $2^x+2^{-x}=5$

Calculation:

Let $2^x=y$ . Then, the given equation becomes

  $\begin{array}{c}y+\frac{1}{y}=5\\ \frac{y^2+1}{y}=5\\ y^2+1=5y\\ y^2-5y+1=0\end{array}$

Now by using the quadratic formula, it follows that

  $\begin{array}{l}y=\frac{-\left(-5\right)\pm \sqrt{{\left(-5\right)}^2-4\left(1\right)\left(1\right)}}{2\left(1\right)}\\ y=\frac{5\pm \sqrt{21}}{2}\end{array}$

Therefore,

  $\begin{array}{l}{2}^x=\frac{5\pm \sqrt{21}}{2}\\ x\log 2=\log \frac{5\pm \sqrt{21}}{2}\\ x=\frac{\log \left(\frac{5\pm \sqrt{21}}{2}\right)}{\log 2}\\ x=\frac{\log \left(\frac{5+\sqrt{21}}{2}\right)}{\log 2}\text{or}x=\frac{\log \left(\frac{5-\sqrt{21}}{2}\right)}{\log 2}\\ x=+2.26\text{or}x=-2.26\end{array}$

Check:

The graph is

From the graph, it can be seen that the points of intersection are $y=2^x+2^{-x}$ and $y=5$ are $\left(-2.26,5\right)$ and $\left(2.26,5\right)$ .

Therefore, the solutions are $x=-2.26,2.26$ .