Category Archives: Sullivan Chapter 6

5.1 Exponential Functions, 5.4 Exponential and Logarithmic Equations, 6.6 Logarithmic and Exponential Equations

Solve an Exponential Equation: Relating the Bases

Problem:  Solve the exponential equation.

36^{x}=(1/6)^{x-3}

Solution:

 

The exponential equation
36^{x}=(1/6)^{x-3}
The bases of the exponents on each side of the equation can be made the same.  36 can be written as 6^2 and 1/6 can be written as 6^{-1}
36^{x}=(1/6)^{x-3}
(6^2)^{x}=(6^{-1})^{x-3}

Use the power rule for exponents to multiply the exponents.
Power Rule for Exponents
(b^m)^n=b^{mn}
(6^2)^{x}=(6^{-1})^{x-3}
6^{2x}=6^{-1(x-3)}
6^{2x}=6^{-x+3}
Exponential functions are one-to-one thus giving us the property that if the bases are the same the exponents are equal.
If b^x=b^y then x=y
6^{2x}=6^{-x+3}
2x=-x+3
Solve the remaining equation.  This equation is linear, first get the variables to the same side.
2x=-x+3
2x+x=-x+x+3
3x=3 
Solve:  Get the variable by itself.
3x=3
{3x}/3={3}/3
x=1
Check: Substitute into the original equation.
36^{1}=(1/6)^{1-3}
36=(1/6)^{-2}
36=6^{2}
36=36

The solution to the equation 36^{x}=(1/6)^{x-3} is x=1.

Here is a video example of a similar type of problem.

 

5.3 Properties of Logarithms, 6.5 Properties of Logarithms

Condense a Logarithm

Problem: Use the properties of logarithms to rewrite the expression as a single logarithm.  Whenever possible, evaluate logarithmic expressions.

ln sqrt{x}- 1/6 ln x +ln root{4}{x}

Soluton:

Logarithmic Expression
ln sqrt{x}- 1/6 ln x +ln root{4}{x}
Use the power rule for logarithms. 
log_b M^p=plog_b M
The coefficient of 1/6 on the middle term becomes the power on the expression inside the logarithm
ln sqrt{x}- 1/6 ln x +ln root{4}{x}
ln sqrt{x}-  ln x^{1/6} +ln root{4}{x}
A radical can be written as a fractional power.  A square root is the same as the one-half power.  A fourth root is the same as the one-fourth power
ln sqrt{x}-  ln x^{1/6} +ln root{4}{x}
ln x^{1/2}-  ln x^{1/6} +ln x^{1/4}
Condense the logarithms using the product and quotient rule.
Product Rule for Logarithms: log_b (MN)=log_b M+log_b N
Quotient Rule for Logarithms: log_b (M/N)=log_b M-log_b N
The expressions inside the logarithm will be positioned in the numerator if the logarithm is positive or will be positioned in the denominator if the logarithm is negative.
ln x^{1/2}-  ln x^{1/6} +ln x^{1/4}
ln {x^{1/2}x^{1/4}}/x^{1/6}
Since these base of the exponential expressions are the same, combine using the power and quotient rules for exponent.
Product Rule for Exponents:
b^Mb^N=b^(M+N)
Quotient Rule for Exponents:
b^M/b^N=b^(M-N)
ln {x^{1/2}x^{1/4}}/x^{1/6}
ln {x^(1/2+1/4)}/x^{1/6}
ln x^(1/2+1/4-1/6)
Find a common denominator to combine the fractions.
ln x^(1/2+1/4-1/6)
ln x^(6/12+3/12-2/12)
ln x^{7/12}

 

Here is a video with a similar example worked out.

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.

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))