Category Archives: MAC1140

5.3 Complex Zeros; Fundamental Theorem of Algebra

Form a Polynomial given the Degree and Zeros

Example: Form a polynomial f(x) with real coefficients having the given degree and zeros.

Degree 4; Zeros -2-3i; 5 multiplicity 2

Solution:

By the Fundamental Theorem of Algebra, since the degree of the polynomial is 4 the polynomial has 4 zeros if you count multiplicity.

There are three given zeros of -2-3i, 5, 5.

The remaining zero can be found using the Conjugate Pairs Theorem.  f(x) is a polynomial with real coefficients.  Since -2-3i is a complex zero of f(x) the conjugate pair of -2+3i is also a zero of f(x).

Now that all the zeros of f(x) are known the polynomial can be formed with the factors that are associated with each zero.

Since f(x) has a zero of 5, f(x) has a factor of x-5

Since f(x) has a second zero of 5, f(x) has a second factor of x-5

Since f(x) has a factor of -2-3i, f(x) has a factor of x-(-2-3i)

Since f(x) has a factor of -2+3i, f(x) has a factor of x-(-2+3i)

Form the polynomial using all of the factors.  The leading coefficient will remain unknown.
f(x)=a(x-5)(x-5)(x-(-2-3i))(x-(-2+3i))
Multiply the factors with complex numbers.  Doing so will cancel the complex numbers from the expression

  • Distribute the minus
  • Multiply each term in one factor by each term in the other factor
  • simplify i^2=-1
  • combine like terms
f(x)=a(x-5)(x-5)(x-(-2-3i))(x-(-2+3i))
f(x)=a(x-5)^2(x+2+3i)(x+2-3i)
f(x)=a(x-5)^2(x^2+2x-3ix+2x+4-6i+3ix+6i-9i^2)
f(x)=a(x-5)^2(x^2+2x-3ix+2x+4-6i+3ix+6i-9(-1))
f(x)=a(x-5)^2(x^2+2x-3ix+2x+4-6i+3ix+6i+9)
f(x)=a(x-5)^2(x^2+4x+13)
Multiply the other pair of factors
f(x)=a(x-5)^2(x^2+4x+13)
f(x)=a(x-5)(x-5)(x^2+4x+13)
f(x)=a(x^2-5x-5x+25)(x^2+4x+13)
f(x)=a(x^2-10x+25)(x^2+4x+13)
Multiply the two trinomials by multiplying each term in the first trinomial by each term in the other trinomial and then combine like terms
f(x)=a(x^2-10x+25)(x^2+4x+13)
f(x)=a(x^4+4x^3+13x^2-10x^3-40x^2-130x+25x^2+100x+325)
f(x)=a(x^4-6x^3-2x^2-30x+325)

The polynomial with degree 4 and zeros of -2-3i and 5 wiht multiplicity 2 is f(x)=a(x^4-6x^3-2x^2-30x+325)

5.5 Applications of Exponential and Logarithmic Functions, 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models

Application: Exponential Growth and Decay (Half-life)

Example: The half life of radium is 1690 years. If 50 grams are present now, how much will be present in 630 years?

Solution:  There is a two part process to this problem.  Part 1: Use some of the information to find the decay rate of radium.  Part 2: Answer the question using the rest of the given information.

Part 1:  Find the decay rate of radium.

Since we are using an exponential model for this problem we should be clear on the parts of the exponential decay model.

Exponential Decay Model

A=A _0 e^{kt}

A _0 is the initial amount

k is the decay rate

t is the time

A is the amount after t time has passed

Since radium has a half life of 1690 years, we know that after 1690 years there will be half of the initial amount of radium left.  This allows me to establish a relationship between the initial amount and the amount after 1690.   The amount after 1690 year is half of the initial amount.   A=1 /2 A _0 when the t=1690.

Substitute these values into the exponential decay formula and solve for k.

 

Exponential Decay Formula
A=A _0 e^{kt}
Make a substitution for A and t since it is known that the half-life is 1690 years A=1 /2 A _0 and t=1690
A=A _0 e^{kt}
1 /2 A _0=A _0 e^{k(1690)}
Solve for the decay rate k:
Start by dividing both sides by the coefficient to isolate the exponential factor
1 /2 A _0=A _0 e^{k(1690)}
{1 /2 A _0}/A_0={A _0 e^{k(1690)}}/A_0
1 /2= e^{k(1690)}
Solve for the decay rate k:
Take the natural log of both sides to get k out of the exponent
1 /2= e^{k(1690)}
ln(1 /2)= ln(e^{k(1690)})
Solve for the decay rate k:
Use the power rule for logarithms to get k out of the exponent
ln(1 /2)= ln(e^{k(1690)})
ln(1 /2)= k(1690)ln(e)
Solve for the decay rate k:
Simplify ln e = 1
ln(1 /2)= k(1690)ln(e)
ln(1 /2)= k(1690)(1)
ln(1 /2)= k(1690)
Solve for the decay rate k:
Solve for k by dividing by 1690 on both sides
ln(1 /2)= k(1690)
{ln(1 /2)}/1690= {k(1690)}/1690
{ln(1 /2)}/1690= k
-0.000410 approx k

Using only the information about radium having a half-life of 1690 years I have found the decay rate for radium.

k={ln(1 /2)}/1690
k approx -0.000410

Note: Although I have put an approximation for k here, try not to round until the very last step.

Part 2:  Answer the question using the rest of the given information.

Given information: If 50 grams are present now, how much will be present in 630 years?

With this information I can identify the initial amount of radium as 50 grams and the time to be 630 years.  Symbolically that is A _0=50 when the t=630.  Substitute the given information and the decay rate k found in part 1 to the exponential decay formula.

Exponential Decay Formula
A=A _0 e^{kt}
Make a substitution for initial amount (A_0), time (t), and the decay rate (k).
A _0=50 , t=630, and k={ln(1 /2)}/1690
A=A _0 e^{kt}
A=50 e^{630{ln(1 /2)}/1690}
Type the expression into your calculator and round to the thousandth place.
A=50 e^{630{ln(1 /2)}/1690}
A approx 38.615

A approx 38.615

38.615 grams will be present 630 years later is 50 grams are present initially.

Here is a video example that is similar to the above example and it shows how to enter information in the calculator.

8.5 Partial Fraction Decomposition

Partial Fraction Decomposition

Example:  Find the partial fraction decomposition for the rational expression.

11/{x(x^2+11)}

Solution:

The rational expression
 11/{x(x^2+11)}
Factor the denominator completely
11/{x(x^2+11)}
Write the expression with each factor as a separate fraction.
11/{x(x^2+11)}=?/x+?/{x^2+11}
Decide what type of expression to put in the numerator

x is linear so we put a constant in the numerator

x^2+11 is a quadratic so we put a linear expression in the numerator
11/{x(x^2+11)}=?/x+?/{x^2+11}
11/{x(x^2+11)}=A/x+?/{x^2+11}
11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}
Now that the set up is done we need to solve for the unknowns A, B and C
11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}
Multiply both sides of the equation by the common denominator {x(x^2+11)} and simplify by canceling out common factors
11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}
{x(x^2+11)}11/{x(x^2+11)}={x(x^2+11)}(A/x+{Bx+C}/{x^2+11})
11={x(x^2+11)}{A/x}+{x(x^2+11)}{Bx+C}/{x^2+11}
11={(x^2+11)}{A}+x(Bx+C)
Use the distributive property and collect the like terms with the x’s
11={(x^2+11)}{A}+x(Bx+C)
11=Ax^2+11A+Bx^2+Cx
11=Ax^2+Bx^2+Cx+11A
11=(A+B)x^2+Cx+11A
Equate coefficients to create a system of equations to solve
11=(A+B)x^2+Cx+11A
0x^2+0x+11=(A+B)x^2+Cx+11A
On the left 11 is the constant term.  There is no linear term so the coefficient is zero and there is no quadratic term so the coefficient is zero.

On the right the constant (no x’s) is 11A, the linear coefficient is C and the quadratic coefficient is A+B.
System of equations from equating the coefficients
matrix{3}{1}{{11=11A} {0=C} {0=A+B}}

Now solve the system of equations.

The first equation is  11=11A.  Since it only has one variable, I can solve for A by hand.

11=11A
11/11={11A}/11
1=A

The second equation is 0=C .  It is already solved.

The third equation is 0=A+B.  Since I know A=1, I can substitute and solve for B.

0=A+B
0=1+B
0-1=1-1+B
-1=0+B
-1=B

Using the original set up of 11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}, substitute the values of A, B, and C.

11/{x(x^2+11)}=A/x+{Bx+C}/{x^2+11}
11/{x(x^2+11)}=1/x+{-1x+0}/{x^2+11}
11/{x(x^2+11)}=1/x+{-x}/{x^2+11}

The partial fraction decomposition of 11/{x(x^2+11)} is 1/x+{-x}/{x^2+11}

5.5 Applications of Exponential and Logarithmic Functions, 6.7 Financial Models

Application of Exponential Functions: Finding the Interest Rate

Example:

What is the interest rate necessary for an investment to quadruple after 7 year of continuous compound interest?

Solution:

Since this question involve continuous compound interest, we will use the associated formula.

A=Pe^{rt}

We are given that the invest quadruples in 7 years.  This tells me that when t=7 that A will be 4 times P.  I can write that in symbols A=4P.

Substitute these values into the continuous compound formula and solve for the interest rate.

Continuous compound formula
A=Pe^{rt}
Substitute the values of t and A into the formula
A=Pe^{rt}
4P=Pe^{r*7}
4P=Pe^{7r}
Solve for r by dividing both sides by P and simplifying
4P=Pe^{7r}
{4P}/P={Pe^{7r}}/P
4=e^{7r}
Solve for r by taking the log of both sides.
4=e^{7r}
ln 4=ln e^{7r}
Solve for r by using the power rule and simplifying
ln 4=ln e^{7r}
ln 4=7r ln e
ln 4=7r (1)
ln 4=7r
Solve for r by dividing both sides by 7 and simplifying
ln 4=7r
{ln 4}/7={7r}/7
{ln 4}/7=r
Find the value in the calculator
{ln 4}/7=r
0.1980420516=r
Write the answer as a percentage rounded to two decimal places
r=0.1980420516
r=19.80420516%
r=19.80%

 

5.4 Exponential and Logarithmic Equations, 6.6 Logarithmic and Exponential Equations, Exponential Equations

Solve an exponential equation: Take the log of both sides

Example:

(1.41)^x = (sqrt{2})^{1-4x}

Solution:

 

The exponential equation
 (1.41)^x = (sqrt{2})^{1-4x}
Since the bases cannot be easily written the same use the method of taking the log of both sides
(1.41)^x = (sqrt{2})^{1-4x}
ln (1.41)^x = ln (sqrt{2})^{1-4x}
Use the power rule for logarithms
ln (1.41)^x = ln (sqrt{2})^{1-4x}
x ln (1.41) = (1-4x) ln sqrt{2}
Use the distributive law
x ln (1.41) = (1-4x) ln sqrt{2}
x ln (1.41) = ln sqrt{2}-4x ln sqrt{2}
Collect the terms with x to one side and collect the terms without x on the other side
x ln (1.41) = ln sqrt{2}-4x ln sqrt{2}
x ln (1.41) +4x ln sqrt{2}= ln sqrt{2}-4x ln sqrt{2} +4x ln sqrt{2}
x ln (1.41) +4x ln sqrt{2}= ln sqrt{2}
Factor the common x
x ln (1.41) +4x ln sqrt{2}= ln sqrt{2}
x( ln (1.41) +4 ln sqrt{2})= ln sqrt{2}
Solve for x by dividing both sides by the factor in the parenthesis and simplify
x( ln (1.41) +4 ln sqrt{2})= ln sqrt{2}
{x( ln (1.41) +4 ln sqrt{2})}/{ ln (1.41) +4 ln sqrt{2}}= {ln sqrt{2}}/{ ln (1.41) +4 ln sqrt{2}}
x = {ln sqrt{2}}/{ ln (1.41) +4 ln sqrt{2}}
The solution
x = {ln sqrt{2}}/{ ln (1.41) +4 ln sqrt{2}}
x =0.2003

When you type this into a calculator be sure to use parenthesis around the numerator and around the denominator.  Here is an example of how you might enter it.

(ln (sqrt{2}))/(ln(1.41)+4 ln(sqrt{2}))

Here is a youtube video with a similar example.

5.3 Properties of Logarithms, 5.4 Exponential and Logarithmic Equations, 6.6 Logarithmic and Exponential Equations, Logarithmic Equations

Solve the Logarithmic Equation by the one to one property

Example:

2 log_3(7-x)-log_3 2=log_3 18

Solution:

 

The logarithmic equation
2 log_3(7-x)-log_3 2=log_3 18
Use the power rule and the quotient rule to condense to a single logarithm
2 log_3(7-x)-log_3 2=log_3 18
log_3(7-x)^2-log_3 2=log_3 18
log_3((7-x)^2/ 2)=log_3 18
Since both sides of the equation have the same log base the expressions inside the logarithms must be equal
log_3((7-x)^2/ 2)=log_3 18
(7-x)^2/ 2= 18
Clear the denominator by multiplying by 2 on both sides and simplifying
(7-x)^2/ 2= 18
2*(7-x)^2/ 2= 2*18
(7-x)^2= 36
Get rid of the square by square rooting both sides and simplifying
(7-x)^2= 36
sqrt{(7-x)^2}= sqrt{36}
7-x= pm 6
Get x by itself by subtracting 7 on both sides
7-x= pm 6
7-7-x=-7 pm 6
-x=-7 pm 6
Get x by itself by dividing both sides by negative 1
-x=-7 pm 6
-x/-1={-7 pm 6}/-1
x=7 pm 6
x=7 + 6 or x=7 - 6
x=13 or x=1
Check x=13
2 log_3(7-13)-log_3 2=log_3 18
2 log_3(-6)-log_3 2=log_3 18
Log of a negative is undefined.  Exclude this solution.
Check x=1
2 log_3(7-1)-log_3 2=log_3 18
2 log_3(6)-log_3 2=log_3 18
log_3(6)^2-log_3 2=log_3 18
log_3 36-log_3 2=log_3 18
log_3 36/2=log_3 18
log_3 18=log_3 18
Keep this solution.

The solution to the equation is x=1.

Here is a youtube video that is similar.

5.4 Exponential and Logarithmic Equations, 6.6 Logarithmic and Exponential Equations

Solve an Exponential Equation: Take the log of both sides

Problem:  Solve the exponential equation.

16^{3x-3}=3^{x-3}

Solution:

The exponential equation
16^{3x-3}=3^{x-3}
Since the bases cannot be easily written the same, use the method of taking the log of both sides
ln (16^{3x-3})=ln (3^{x-3})
Use the power rule for logarithms.
(3x-3)ln16=(x-3)ln3
Use the distributive property.
3xln16-3ln16=xln3-3ln3
Collect the terms with x to one side and collect the terms without x on the other side.
3xln16-3ln16=xln3-3ln3
3xln16-3ln16+3ln16=xln3-3ln3+3ln16
3xln16-xln3=xln3-xln3-3ln3+3ln16
3xln16-xln3=-3ln3+3ln16
3xln16-xln3=3ln16-3ln3
Factor the common x.
3xln16-xln3=3ln16-3ln3
x(3ln16-ln3)=3ln16-3ln3
Solve for x by dividing both sides by the factor in the parenthesis and simplify.
x(3ln16-ln3)=3ln16-3ln3
{x(3ln16-ln3)}/{3ln16-ln3}={3ln16-3ln3}/{3ln16-ln3}
x={3ln16-3ln3}/{3ln16-ln3}
The solution
x={3ln16-3ln3}/{3ln16-ln3}
x = 0.6956

When you type this into a calculator be sure to use parenthesis around the numerator and around the denominator.  Here is an example of how you might enter it.

(3ln(16)-3ln(3))/(3ln(16)-ln(3))