Test 19 (Lessons 37–38): Exponential Functions, Equations, and Inequalities Solutions

For problems 1–2, use the equations: $f (x) = 2 {(0.3)}^{x - 1} + 5$ and $g (x) = - 2 {(0.3)}^{x + 5} - 3$

Describe the transformation from f(x) to g(x).

The function g(x) is reflected over the x-axis. Then it shifts left 6 units and down 8 units as compared to f(x).

	f(x)	g(x)
a	2	–2
h	1	–5
k	5	–3

Explain whether f(x) and g(x) represent growth or decay. Name the domain and range.

Both functions are decay, because $b = 0.3,$ and this is between zero and one.

${domain}_{f} : \{x | x \in ℝ\} {domain}_{g} : \{x | x \in ℝ\} {range}_{f} : \{y | y \in ℝ, y > 5\} {range}_{g} : \{y | y \in ℝ, y < - 3\}$

Write the exponential equation in the form $y = a b^{x}$ using the points $(- 4, 48)$ and $(2, 0.75) .$

$\begin{matrix} y = a b^{x} \\ 48 = a b^{- 4} & 0.75 = (48 b^{4}) b^{2} \\ 48 = \frac{a}{b^{4}} & \frac{0.75}{48} = \frac{48 b^{6}}{48} \\ a = 48 b^{4} & \sqrt[6]{\frac{1}{64}} = \sqrt[6]{b^{6}} \\ b = \frac{1}{2} \end{matrix}$

$\begin{matrix} a = 48 {(\frac{1}{2})}^{4} \\ a = 48 (\frac{1}{16}) \\ a = 3 & y = 3 {(\frac{1}{2})}^{x} \end{matrix}$

Solve.

$8^{9 - 4 x} > 4^{9 x + 3}$

$\begin{array}{rcl} 8 = 2^{3} 4 = 2^{2} \\ 2^{3 (9 - 4 x)} & > & 2^{2 (9 x + 3)} \\ 3 (9 - 4 x) & > & 2 (9 x + 3) \\ 27 - 12 x & > & 18 x + 6 \\ \frac{21}{30} & > & \frac{30 x}{30} \\ x & < & \frac{7}{10} \end{array}$

$729^{2 x - 3} = {(\frac{1}{27})}^{x - 9}$

$\begin{array}{rcl} 729 = 3^{6} \frac{1}{27} = 3^{- 3} \\ 3^{6 (2 x - 3)} & = & 3^{- 3 (x - 9)} \\ 6 (2 x - 3) & = & - 3 (x - 9) \\ 12 x - 18 & = & - 3 x + 27 \\ 15 x & = & 45 \\ x & = & 3 \end{array}$

For problems 6–7, use the exponential equation: $y = 4^{x}$

Name b and explain if it represents the growth or decay factor.

b = 4
It is a growth factor because b > 1.

Graph.

x	y
–1	0.25
0	1
1	4

Solve.

$6^{x} \leq 36^{x - 3}$

$\begin{array}{rcl} 36 & = & 6^{2} \\ 6^{x} & \leq & 6^{2 (x - 3)} \\ x & \leq & 2 (x - 3) \\ x & \leq & 2 x - 6 \\ 6 & \leq & x \\ x & \geq & 6 \end{array}$

$25^{x - 3} = 125^{\frac{1}{2} x - 1}$

$\begin{array}{rcl} 25 = 5^{2} 125 = 5^{3} \\ 5^{2 (x - 3)} & = & 5^{3 (\frac{1}{2} x - 1)} \\ 2 (x - 3) & = & 3 (\frac{1}{2} x - 1) \\ 2 x - 6 & = & \frac{3}{2} x - 3 \\ \frac{1}{2} x & = & 3 \\ x & = & 6 \end{array}$

${(\frac{8}{125})}^{2 x} = {(\frac{25}{4})}^{1 - 5 x}$

$\begin{array}{rcl} (\frac{8}{125}) = {(\frac{2}{5})}^{3} (\frac{25}{4}) = {(\frac{2}{5})}^{- 2} \\ {(\frac{2}{5})}^{3 (2 x)} & = & {(\frac{2}{5})}^{- 2 (1 - 5 x)} \\ 3 (2 x) & = & - 2 (1 - 5 x) \\ 6 x & = & - 2 + 10 x \\ - 4 x & = & - 2 \\ x & = & \frac{1}{2} \end{array}$