6.1 Exponential Exercises

The value of a car is V(x)= 10(.75)^x, 
where x>=0 is the age of the car in years and 
V(x) is its value in thousands of $.

Evaluate:
V(0)   = 10 or $10,000
V(1)   = 7.5   
V(2)   = 5.625 or $5,625
V(3)   = 4.21875
V(5)   = 2.373047
V(10)  = 0.563135
V(20)  = 0.031712  or $31.71

Solve:
V(x)=10   x= 0
V(x)=5    x= 2.409421
V(x)=2.5  x= 4.818842
V(x)=1.25 x= 7.228263
Based on these, how much time does it take for the value of the car to halve?
   ~2.41 years
Graph the function.

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The price of a stock of MEGA Corp. is P(x)= 10(1.25)^x, 
where x>=0 is the numbers of years from today and 
P(x) is the stock's price in $.

Evaluate:
P(0)    = 10
P(1)    = 12.5
P(2)    = 15.625
P(3)    = 19.53125
P(5)    = 30.517578
P(10)   = 93.132257
P(20)   = 867.361738

Solve:
P(x)=10    x= 0
P(x)=20    x= 3.106284
P(x)=40    x= 6.212567
P(x)=80    x= 9.318851
Based on these, how much time does it take for the price of the stock to double?
   ~3.11 years
Graph the function.

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Some 60°C warm stuff is put into a 10°C cool place. Its temperature T after t minutes
is T(t)= 10 + 50e^(-0.1t)

Evaluate:
T(0)    = 60
T(1)    = 55.241871
T(2)    = 50.936538
T(5)    = 40.326533
T(10)   = 28.393972
T(20)   = 16.766764
T(30)   = 12.489353
T(50)   = 10.336897

What looks to be the horizontal asymptote as t gets large?
  10 C

Solve:
T(t)=60   x= 0
T(t)=50   x= 2.23144
T(t)=40   x= 5.10826
T(t)=30   x= 9.16291
T(t)=20   x= 16.09438
T(t)=15   x= 23.02585
T(t)=11   x= 39.12023
T(t)=10.5 x= 46.0517
T(t)=10.1 x= 62.14608
T(t)=10.01 x= 85.17193
T(t)=10    x goes to infinity

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World population doubled in the 40 years between 1960 and 2000, 
from 3 billion to 6 billion.
From this fact, a function that predicts future world population is:
  WP(t)= 6*2^(0.025t)
where t is years since 2000. 

Evaluate:
If this trend continues (it hasn't but pretend it did),
what should the world population be 
in 2021?    8.633601
in 2000?    6
in 2050?    14.270485
in 2100?    33.941125

Solve:
When will the world population be 
 10 billion? WP(t)=10     t= 29.47864   2029
  6 billion? WP(t)=6      t= 0          2000
100 billion? WP(t)=100    t= 162.35576  2162

Graph the function.

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Carbon-14 is radioactive, which means some random atoms of a pile
of it will change to some other kind of atom at a known rate. 
Half of it will have changed in 5730 years (its "half-life").
The decimal percent amount of the starting pile that Still is carbon-14
after t years is:
  S(t)= e^(-0.000121t)
                       
Evaluate:
How much is still carbon-14 after 
 0 years: S(0)              = 1  = 100%
 1000 years: S(1000)        = 0.886034 = 88.6034%
 2000 years: S(2000)        = 0.785056 = 78.5056%
 10000 years: S(10000)      = 0.298197 = 29.8197%
 20000 years: S(20000)      = 0.088922 =  8.8922%
 100000 years: S(100000)    = 0.000006 =  0.0006%

Solve:
How many years for there to be one half of the starting amount: S(t)=.5       t= 5728.487603
How many years for there to be one quarter of the starting amount: S(t)=.25   t= 11456.975207
How many years for there to be one eighth of the starting amount: S(t)=.125   t= 17185.46281
How many years for there to be one sixteenth of the starting amount: S(t)=.0625  t= 22913.958678

Graph the function.