7.7 Write and Apply Exponential and Power Functions

Example 1

Write an exponential function y = ab^x whose graph passes through (1, 10) and (4, 80).

Solution

1. Substitute the coordinates of the two given points into y = ab^x.

__10__ = ab^1	Substitute 10 for y and 1 for x.
__80__ = ab^4	Substitute 80 for y and 4 for x.

2. Solve for a in the first equation to obtain a = and substitute this expression for a into the second equation.

80	=		Substitute for a.
80	=	_10b^3_	Simplify.
_8_	=	_b^3_	Divide each side by _10_ .
_2_	=	_b_	Take the positive cube root.

3. Because b = _2_ , it follows that a = So, y = __5 · 2^x___ .

Example 2

Savings The table shows the amount A in a savings account t years after the account was opened.

t	0	1	2	3	4	5	6	7
A	210	255	310	377	459	557	677	822

· Draw a scatter plot of the data pairs (t, ln A). Is an exponential model a good fit for the original data pairs (t, A)?

· Find an exponential model for the original data.

Solution

1. Use a calculator to create a table of data pairs (t, In A).

t	0	1	2	3	4
ln A	_5.35_	_5.54_	_5.74_	_5.93_	_6.13_

t	5	6	7
ln A	_6.32_	_6.52_	_6.71_

2. Plot the new points. The points lie close to a line, so an exponential model should be a good fit for the original data.

3. To find an exponential model A = ab^t, choose two points on the line, such as
(1, _5.54_ ) and (6, _6.52_ ). Use these points to find an equation of the line. Then solve for A.

Because the axes are t and In A, the point-slope form is rewritten as
In A - In
A₁ = m(t - t₁).

In A - _5.54__	=	_0.196_ (t - 1)	Equation of line
ln A	=	_0.196t + 5.344_	Simplify
A	=	_e^0.196t + 5.344_	Exponentiate each side using base e.
A	=	__e^5.344(e^0.196)t__	Use properties of exponents.
A	=	__209.35(1.22)t__	Exponential model

Example 3

Write a power function y = ax^b whose graph passes through (2, 4) and (6, 10).

1. Substitute the coordinates of the two given points into y = ax^b.

4	=	_a · 2^b_	Substitute 4 for y and 2 for x.
10	=	_a · 6^b_	Substitute 10 for y and 6 for x.

2. Solve for a in the first equation to obtain a = and substitute for a into the second equation.

10	=		Substitute for a.
10	=	__4 · 3^b__	Simplify.
_2.5_	=	__3^b__	Divide each side by _4_ .
__log3 2.5__	=	__b__	Take __log3__ of each side.
	=	__b__	Change-of-base formula
_0.83_	=	__b__	Use a calculator.

3. Because b = _0.83_ , it follows that a = So, y = __2.25x^0.83___ .

Example 4

The table gives the approximate volume V of spheres with radius r. Draw a scatter plot of the data pairs (ln r, ln V). Is a power model a good fit for the original data pairs (r, V)? Find a power model for the original data.

r	1	2	3	4	5	6
V	4.189	33.510	113.097	268.083	523.599	904.779

1. Use a calculator to create a table of data pairs (ln r, ln V).

ln r	0	0.693	1.099	1.386	1.609	1.792
ln V	1.432	3.512	4.728	5.591	6.261	6.808

2. Plot the new points. The points appear linear, so a power model should be a good fit for the original data.

3. To find a power model V = ar^b, choose two points on the line, such as
(1.099, __4.728__ ) and (1.792, __6.808__ ). Use these points to find an equation of the line. Then solve for V.

In V - _4.728__	=	_3_ (ln r - __1.099__ )	Equation of line
In V	=	__ln r^3 + 1.431__	Power property of logarithms
V	=	__e ^ln^r3 + 1.431__	Exponentiate each side.
V	=	__e ^ln^r3 · e^1.431__	Product of powers property
V	=	__4.183r^3__	Simplify.

Because the axes are ln r and In V, the point-slope form is rewritten as In V - In V₁ =
m(ln r - ln r₁).