#$&*
course Mth 173
10/7 17:20
Text_08Problem 1.4.10
Given 4 * 3^x = 7 * 5^x
Rearranging the equations we get
(3/5) ^ x = 7 / 4
0.6 ^ x = 1.75
Thus x = log{base 0.6}(1.75) = log(1.75) / log(0.6) = -1.0955
Thus the solution of the given equation is x = -1.0955
Problem 1.4.12
Given equation 2^x = e ^ (x + 1)
Since a ^ (b + c) = a ^ b * a ^ c
Thus 2 ^ x = e ^ x * e
Rearranging the equation we get (2 / e) ^ x = e, (0.736) ^ x = e
Thus taking the log rules x = log{base 0.736} (e) = log(e) / log(0.736)
Approximating the value further with the value of e = log(2.718) / log(0.736) = -3.262 approx.
Thus the solution of the given function is -3.262 approx.
Problem 1.4.16
Given function 10 ^ (x + 3) = 5 * e ^ (7 - x)
Using the properties of log as mentioned in the earlier question,
10^3 * 10^x = 5 * [e^7 / e^x]
200 * 10^x = e^7 / e^x
(10*e)^x = e^7 / 200
Thus x = log{ base 10e }( e^7 / 200 )
Approximating the value of x = log{base 27.183}(5.483) = log(5.483)/log(27.183) = 0.515
Thus the approximate value of the function is 0.515
Problem 1.4.19
Given expression a = b^t
Since a and b are both positive values
Using log rules we get t = log{ base b}(a) = log(a) / log(b)
Problem 1.4.23
Given expression a = b*(e^t)
Since a and b are both positive values
Rearranging the expression (a/b) = e^t
Using log rules we get t = log{ base e }( a/b ) = log(a/b) / log(e) = ( log(a) - log(b) ) / log(e)
= ln(a/b) = ln(a) - ln(b)
Problem 1.4.26
Given expression P = 10 * ( 1.7 ^ t )
We need to compare this expression to the standard expression P = P0 * (e ^ kt)
Comparing the expression we get P0 = 10 and e ^ kt = 1.7 ^ t
Using the log rules we also know that e ^ kt = (e ^ k) ^ t = 1.7 ^ t
Thus e ^ k = 1.7
k = log{ base e }(1.7) = log ( 1.7 ) / log(e) = approximately 0.5306
thus the given expression can be written as P = P0 * ( e ^ kt ) where P0 = 10 and k = 0.5306 approximately.
PROBLEM 1.4.27
Given expression P = 174 * ( 0.9 ) ^ t
We need to compare this expression to the standard expression P = P0 * ( e ^ kt )
Comparing the expression we get P0 = 174 and e ^ kt = 0.9 ^ t
Using the log rules we also know that e ^ kt = ( e ^ k ) ^ t = 0.9 ^ t
Thus e ^ k = 0.9
k = log{ base e }(0.9) = log ( 0.9 ) / log(e) = approximately - 0.1054
thus the given expression can be written as P = P0 * ( e ^ kt ) where P0 = 174 and k = -0.1054 approximately.
Problem 1.4.29
Given function p(t) = (1.04)^t
Let p(t) be mentioned as y and t as x , then we have
y = (1.04) ^ x
In order to find the inverse we need to find the function x in terms of y , thus we have
x = log{ base 1.04 }(y) = log(y) / log(1.04)
now to make the dependent variable independent and vice versa x and y are interchanged. Thus we have
y = log(x) / log(1.04) , where y = p^-1(t) and x = t
thus approximately p^-1(t) = 58.708*log(t)
Problem 1.4.32
Given population in 2000 = 40,000,000 and population in 2010 = 48,000,000
Let t be the number of years since 2000.
Thus number of years of 2010 since 2000 = 10 years.
Let the 2 points be plotted on the graph we will have ( 0 , 40000000 ) and ( 10 , 48000000 )
Let the exponential graph be y = a * b ^ x , where y is the population and x is the number of years since 2000.
40000000 = a * b ^ 0 , thus 40000000 = a
And using the 2nd point, 48000000 = 40000000 * b ^ 10
Thus 1.2 = b ^ 10
Thus b = 1.2 ^ .1 = 1.0184 approximately.
Thus the given exponential function is y = 40000000 * ( 1.0184 ^ x )
Population predicted in 2020, t = 20 years thus y = 40000000 * ( 1.0184 ^ 20 ) = 57600705.7
For doubling time we need to find t for population of 80000000 , thus
80000000 = 40000000 * ( 1.0184 ^ t )
2 = 1.0184 ^ t
Using log rules we get
t = log{ base 1.0184 }(2) = log(2) / log(1.0184) = 38.0166 approximately.
Thus the doubling time for the population is 38.0166 years.
Problem 1.4.34
We know that the number of bacteria originally is 500 and after 2 hours it grows to reach 1500.
Thus the 2 points when plotted on the graph are ( 0 , 500 ) and ( 2 , 1500 )
Since the graph is an exponential graph thus it satisfies the basic exponential function y = a * b ^ x where y is the number of bacteria and x is the number of hours.
Since both the points satisfy the equation we can find the values of the arbitrary constant by substituting the points in the equation. Thus using the 1st point we get
500 = a * b ^ 0 = a * 1
Thus a = 500
And using the 2nd equation we get
1500 = a * b ^ 2 = 500 * b ^ 2
3 = b ^ 2 , thus b = 1.732 approx
Thus the function obtained is y = 500 * ( 1.732 ^ x )
Thus the number of bacteria in 6 hours will be
y = 500 * (1.732 ^ 6) = 500 * 26.995 = 13497.5 bacteria
Problem 1.4.39
The expression given for remaining drug in the body is A = 10 * (0.82) ^ t
(a) For the initial amount taken we see the drug remaining in the body at time t = 0, that is the initial time thus A (0) = 10 * (0.82) ^ 0 = 10 * 1 = 10 mg
Thus the initial amount of drug taken is 10 mg
(b) First we find the drug remaining in the body after 1 hour
Thus A (1) = 10 * (0.82) ^ 1 = 8.2 mg
Thus the amount of drug that left the body in an hour = 10 mg - 8.2 mg = 1.8 mg
Thus the percentage of drug leaving the body in 1 hour = (1.8 / 10) * 100 = 18 %
And since for an exponential function the ratio of successive values of y is the same thus this percentage of change of drug from 0 to 1 hour, or 1 hour to 2 hour or further is going to remain the same.
(c) The remaining drug after 6 hours will be
A (6) = 10 * (0.82) ^ 6 = 3.04 mg
Thus 3.04 mg of drug is administered 6 hours after the dose is taken.
(d) now find the time for 1 mg of drug remaining in the body
Thus 1 = 10 * ( 0.82) ^ t
Rearranging using log rules
0.1 = 0.82 ^ t
t = log{ base 0.82 }(0.1) = log ( 0.1) / log ( 0.82 ) = 11.6028 approximate hours.
Thus after 11.6028 hours the amount of drug remaining in the body will be 1 mg.
Problem 1.4.40
Given initial amount of caffeine in coffee = 100 mg
(a) Rate % at which caffeine is increasing = - 17%
Thus the growth rate = - 0.17
And the growth factor = 1 - 0.17 = 0. 83
Let A be the amount of the caffeine in the coffee after t hours
Initial amount = A0 = 100 mg
Thus A = A0 * (growth factor) ^ t = 100 * (0.83 ^ t) mg
Thus the amount of caffeine in the coffee will be A = 100 * (0.83 ^ t) mg
(b) The graph of the exponential function is plotted by using 4 points. The points are (0,100) , (1,83) , (2,68.9) , (3,57.2) and (4,47.5). Using these points a curve is drawn. For the curve the time in hours is on the x axis and the amount in mg of caffeine is on the y axis. Now the initial amount of caffeine is 100 mg, thus the half-life amount is 50 mg of caffeine. Now in order to find the time in hours corresponding to the amount of caffeine, we locate the point (0, 50) on the y axis. A horizontal line parallel to the x aixs is drawn to the curve. From the point of intersection a vertical line is drawn to the x axis to meet at a point. The point of intersection gives us the half-life of caffeine. The value that we obtain by performing the operations on the graph is 3.65 hours. Thus according to my graph caffeine in the body would take 3.65 hours to reach its half value.
(c) Using logarithms to find the half-life. Since the initial amount is 100 mg the half-life amount would be 50 mg. For t corresponding to 50 mg
50 = 100 * (0.83 ^ t)
0.5 = 0.83 ^ t
Using log rules
t = log{base 0.83} (0.5) = log(0.5) / log(0.83) = 3.720 approx half-life time.
Problem 1.4.44
Given the tiger population in 1900 is 100,000 and the population in 2010 is 3200.
(a) Given that the tiger population is decreasing exponentially.
Let t be the number of years since 1900. Thus 2010 is 110 years since 1900. If the points are plotted on the graph, the points are (0, 100,000) and (110, 3200) where the x axis is the time since 1900 axis and y axis is the population.
Let standard exponential function be y = a * (b ^ x) where y is the tiger population and x is the number of years since 1900
And the 2 points satisfy the function,
100,000 = a * (b ^ 0) = a * 1 = a
And
3200 = a * (b ^ 110) where a = 100,000
Thus 0.032 = b ^ 110
Thus b = 0.032 ^ (1 / 110) = 0.9692 approx
Thus the function is y = f(t) = 100,000 * (0.9692 ^ t)
(b) The population of tiger in 2000.
t for 2000 will be = 2000 - 1900 = 100 years
Thus population of tiger in the year 2000 f(100) = 100,000 * (0.9692 ^ 100) = 4378.65
Population of tiger in the year 2010 = 3200
Thus decrease in population = 4378.65 - 3200 = 1178.65
Thus the percentage decrease in 2010 from 2000 = (1178.65 / 4378.65) * 100 = 26.92 %
The given data shows a change of 40 %, thus according to the function the percentage from the data is larger than the percentage predicted by my answer through the function.
Problem 1.4.45
Given data in 2011
The population of China = $1.34 billion
The population of India = $1.19 billion
The annual growth rate of India = 1.37%
The annual growth rate of China = 0.4%
Thus the growth factor of India = 1 + 0.0137 = 1.0137
The growth factor of China = 1 + 0.004 = 1.004
Let PC be the population of China after t years
Let PI be the population of India after t years
Thus we have PC = $1.34 * (1.004 ^ t) billion and
PI = $1.19 * (1.0137 ^ t) billion
We know that the population of China is greater than that of India, thus Indian population will become greater than China only after t when the population of India becomes equal to that of China.
We find t when PC = PI
$1.34 * (1.004 ^ t) billion = $1.19 * (1.0137 ^ t) billion
1.126 = (1.0137 / 1.004) ^ t = 1.01 ^ t
Thus t = ln(1.126) / ln(1.01) = 11.93 years = 12 years approximate
Thus after 12 years ( that is in the year 2023) the population of India will become greater than that of China if the growth rates remain the same.
Problem 1.4.50
Let P0 be the initial population of the bacterial colony.
Let P be the population of the bacterial colony after t hours.
Based on the data we have for t = 5 hours P = 2P0
The exponential function will be P = P0 * (k ^ t)
2P0 = P0 * (k ^ 5), 2 = k^5
Thus 5 = log(2) / log(k) , log(k) = log(2) / 5
Now in order to find the tripling time we substitute P = 3P0, thus
3P0 = P0 * ( k ^ t3 ), where t3 is the tripling time
Rearranging the equation we get
t3 = log(3) / log(k) , we calculated log(k) = log(2) / 5
thus t3 = [ log(3) / log(2) ] * 5 = 1.585 * 5 = 7.925 hours.
Thus the bacterial colony would take 7.925 hours to triple itself.
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