Chapter 12.2, Problem 37STP

To determine

To find: the integer values of x do the given graphs of $y=20{\left(0.5\right)}^x\text{ and }y=0.5\cdot 4^x$ intersect.

Answer to Problem 37STP

  $\left(1,2\right)$ .

Explanation of Solution

Given:

The given graph of

  $\begin{array}{l}y=20{\left(0.5\right)}^x\\ y=0.5\cdot 4^x\end{array}$

Calculation:

As the given graph of $\begin{array}{l}y=20{\left(0.5\right)}^x\\ y=0.5\cdot 4^x\end{array}$ .

Now, substitutes the value of $x=1,x=2$ in the equation $y=20{\left(0.5\right)}^x$ .Therefore,

  $\begin{array}{l}y=20{\left(0.5\right)}^x\left\{\text{given equation}\right\}\\ y=20{\left(0.5\right)}^1\{\text{substitutes }x=1\}\\ \therefore y=20\left(0.5\right)=10\\ y=20{\left(0.5\right)}^x\left\{\text{given equation}\right\}\\ y=20{\left(0.5\right)}^2\{\text{substitutes }x=2\}\\ \therefore y=20\left(0.25\right)=5\end{array}$

Again, substitutes the value of $x=1,x=2$ in the equation $y=0.5\cdot 4^x$ .Therefore,

  $\begin{array}{l}y=0.5\cdot 4^x\left\{\text{given equation}\right\}\\ y=0.5{\left(4\right)}^1\{\text{substitutes }x=1\}\\ \therefore y=0.5\left(4\right)=2\\ y=0.5\cdot 4^x\left\{\text{given equation}\right\}\\ y=0.5{\left(4\right)}^2\{\text{substitutes }x=2\}\\ \therefore y=0.5\left(16\right)=8\end{array}$

Since when you plug 1 into both equations, the second equation gives the higher value, but when you plug in 2 into both equations the first equation gives the higher value, meaning the two graphs must have intersected at some point between 1 and 2.

Thus, the intersect point is $\left(1,2\right)$