   Chapter 4.6, Problem 22E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding Exponential Models In Exercises 21 and 22, find exponential models y 1 = C e k 1 t  and  y 2 =   C ( 2 ) k 2 t that pass through the two given points. Compare the values of k 1  and  k 2 . Briefly explain your results. ( 0 , 8 ) ,   ( 20 , 1 2 )

To determine

To calculate: The exponential models y1=Cek1t and y2=C(2)k2t for the exponential model passing through the points (0,8),(20,12).

Explanation

Given Information:

The provided points are (0,8),(20,12) and the exponential models are y1=Cek1t and y2=C(2)k2t.

Calculation:

Consider the provided points are (0,8),(20,12) and the exponential models are y1=Cek1t and y2=C(2)k2t.

Consider the exponential model y1=Cek1t.

Consider the points (0,8),(20,12).

Substitute t=0,y1=8 in the exponential model y1=Cek1t,

8=Cek108=Ce0C=8

Substitute t=20,y1=12 in the exponential model y1=8ek1t,

12=8ek120e20k1=116e20k1=116

Take natural log on both the sides,

ln (e20k1)=ln (116)20k1 lne=ln (116)20k1=ln (116)k1=ln (116)20

Solve further,

k1=0.1386

Substitute k1=0.1386 and C=8 in the exponential model y1=Cek1t,

y1=8e0.1386t

Hence, the exponential model is y1=8e0

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