OVERVIEW: If a scatterplot shows a curved pattern, itcan perhaps be conveniently modeled by an

exponentialgrowthor decay function of the formy = ab ^{x}or a

powerfunction of the formy = ax ^{b}In these situations, we can linearize the data by making use oflogarithms. Among the advantages of using logarithms is the fact thatuse of logarithms produces smaller numbers, making graphical displaysmore convenient to construct.

Definition: log_{b}x = y if and only ifb^{y} = x. [x> 0, b > 0 and b is not equal to 1]

Rules for logarithms

1. log(AB) = logA+ logB

2. log(A/B) = logA - logB

3. logA

^{p}= plogA

Note that y = ab^{x} (**exponential function**)

==> logy = loga + logb^{x}

==> logy = loga + xlogb.

This is a __linear relationship__ between the variables x andlogy since loga and log b are constants.

Also, y = ax^{b} (**power function**)

==> logy = loga + logx^{b}

==> logy = loga + blogx.

This is a __linear relationship__ between the variables logxand logy.

Here is a simple example using four points. (Use TI-83 to verifycalculations.)

x | y | logy | logx |

2 | 3 | 0.47712 | 0.30103 |

7 | 41 | 1.6128 | 0.8451 |

10 | 168 | 2.2253 | 1.0000 |

16 | 625 | 2.7959 | 1.2041 |

The points {(x,y)} form a curved pattern.

Fitting a **least squares regression line** to {(x,y)} yields

y (hat) = -186.6715 + 45.2482x

r

^{2}= .8573, r = .9259For x = 12, the predicted y value is y(hat) =-186.6715+45.2482(12) = 356.31.

Fitting an **exponential function** to {(x,y)} yields

y (hat) = (2.1347)(1.4640)

^{x}r

^{2}= .9526, r = .9760For x = 12, the predicted value is y(hat) =(2.1347)(1.4640)

^{12}= 206.93

Fitting a **least squares regression line** to {(x,logy)}, weget

log[y(hat)] = 0.3293 + 0.1655x

r

^{2}= .9526, r = .9760For x = 12, we have logy = 2.3153, and y(hat) =10

^{2.3153}= 206.68

Fitting a **power function** to {(x,y)} yields

y(hat) = (0.4416)x

^{2.5464}r

^{2}= .9843, r = .9921For x = 12, y(hat) = (0.4416)(12)

^{2.5464}= 247.20

Fitting a **least squares regression line** to {(logx,logy)}yields

log[y(hat)] = -0.3550 + (2.5464)logx

r

^{2}= .9843, r = .9921

For x = 12, we have log[y(hat)] = -0.3550 + (2.5464)(log12) =2.3930, and y(hat) = 10^{2.3930} = 247.19

Don't forget **residuals**. These are useful in determining thebest model to fit to a data set.

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