Download Section 2.2 Continuing With Operation/Inverse

Section 2.2 Continuing With Operation/Inverse Pairs
In this section we’ll continue working with the operation inverse pairs
Logarithm
Square
Multiplication
Addition




Exponential
Square Root
Division
Subtraction
We’ll start with the other half of the logarithmic/exponential pair.
2.2.1 Solving Logarithmic Equations
A logarithmic equation has a variable in the argument. For instance, log y  4 is a logarithmic
equation. Usually a logarithmic equation will have one solution. To solve an equation like log y  4 you’ll
need to “undo” the logarithmic operator. Here are the two properties we’ll use.
Property – The Exponential Property of Equality
English: Raising both sides of an equation as the power of the same base results in an
equivalent equation.
Example: log y  4 is equivalent to 10log y  104
Algebra: If s  t then bs  bt .
Property – The Second Inverse Property of Logarithms
English: If a logarithm of base b is written as an exponent of base b the expression
simplifies to the argument of the logarithm.
log 8
Example: 2
2
 8 or eln y  y
Algebra: blogb x  x b, x  0 and b  1
Practice 2.2.1 Solving Logarithmic Equations
Solve using the exponential property of equality and the second inverse property of
logarithms. If necessary round your answer to the hundredths place.
a) log y  100  104
Oper
Inv
log
10 ^
100
100

log y  100  100  104  100

log y  4
10log y  104

y  10,000

Copyright 2014 Scott Storla
Built the two column table.
Used the subtraction property of equality.
Used the exponential property of equality. Remember log is
log10 .
Used the second inverse property of logarithms on the left side
and simplified the right side.
2.2 Continuing With Operation/Inverse Pairs
51
b)
ln t
 15  14.5
6
Oper
ln
6
15
Inv
e^
6
15
ln t
 15  15  14.5  15
6
ln t
 0.5
6
6  ln t 
 6  0.5 
1  6 

Built the two column table.

Used the addition property of equality.

Used the multiplication property of equality.
ln t  3
eln t  e3
t  20.0855
ln  20.0855 
6
 15  14.5
Used the exponential property of equality and the second inverse


property of logarithms. Used e since ln is log e .
Substituted and simplified to check the solution.
14.5  14.5
Homework 2.2 Solve using the exponential property of equality and the second inverse property of
logarithms. If necessary round your answer to the ten-thousands place.
1)
2  logt
2)
ln k  0.12
3)
ln p  24  26
4)
1.6  log n  1.4
5)
1  2log t  2
6)
ln x
 3  4
2
2.2.2 Solving Square Root Equations Using the Squaring Property
A square root equation has a variable in the radicand. For instance,
x  3 is a square root
equation. Usually a square root equation will have one solution. To solve an equation like
x  3 you’ll
need to “undo” the idea of a square root. Here’s the property we’ll use.
Property: The Squaring Property of Equality
English: When you square both sides of an equation the new equation will have all the
solutions of the original equation.
Example: The solutions to x  3 are included in the solution set of
Algebra: All the solutions of
x  b are included in the solutions of
 x
2
 x
2
 32
 b2 . x  0
Although it won’t happen in today’s problems you should be aware that the squaring property may
introduce extraneous (extra) solutions. For example the equation x  3 has solution set 3 . If both sides
are squared the new equation x 2  9 has solution set 3,3 . The new solution set contains the original
solution for x  3 but also includes an extraneous second solution.
After using the square root property we use the inverse property of squaring a square root.
Copyright 2014 Scott Storla
2.2 Continuing With Operation/Inverse Pairs
52
Property – The Inverse Property of Squaring a Square Root
English: For nonnegative radicands taking a square root and squaring are
inverse operations.
Example:
 4
Algebra:
 x
2
4
2
 x, x  0
Practice 2.2.2 Solving Square Root Equations Using the Squaring Property
Solve.
a) 4  2 p  18
Oper
Inv
^2
2
2
18
18
4  18  2 p  18  18
14  2 p
14
2 p

2
2
7 p
 7 2  
p


Built the two column table.

Used the subtraction property of equality.

Used the division property of equality.

Used the squaring property of equality the inverse property on the
right, simplified on the left.

Checked the solution.
2
49  p
4  2 49  18
44
Homework 2.2 Solve.
7)
k 8
11) 0.06 
1
t  1.09
2
8)
82 h
12)
5 p
 9  16
4
9)
15  3 x  15
10)
y
 1  3.4
2
2.2.3 Solving Quadratic Equations Using the Square Root Property
If we square the variable we have a quadratic equation. For instance x 2  4 is an example of a
quadratic equation. It’s common for the solution set of a quadratic equation to have two solutions. Notice
the solution set for x 2  4 has the values 2, 2 .
It seems reasonable that to solve x 2  4 we should take the square root of both sides
x2  4
and simplify to get x  2 . Unfortunately you can see this does not lead to an equivalent equation since
we’ve “lost” the solution 2 . To solve x 2  4 we’ll use the square root property.
Copyright 2014 Scott Storla
2.2 Continuing With Operation/Inverse Pairs
53
Property – The Square Root Property
English: If the square of a variable is equal to a constant then the variable itself
is equal to the square root of the constant or its opposite.
Example: If x 2  4 then x  4 or x   4 .
Algebra: If x 2  a then x  a or x   a . a  0 .
Note: The idea that x equals a or  a is often written x   a .
Practice 2.2.3 Solving Quadratic Equations Using The Square Root Property.
Solve and check.
2
a) y  25
y  25 or y   25

Solved for y using the square root property.
y  5 or y  5

Simplified the solutions.
y 5

How the solutions will often be written.

Checked both solutions. Since both solutions lead to
identities both solutions are probably correct.
52  25
 5 2  25
25  25
25  25
b) 1.5x 2  6.1  11.9
Oper
^2
1.5
6.1
Inv

1.5
6.1

1.5 x 2  6.1  6.1  11.9  6.1
Built the two column table.

Used the addition property of equality.
1.5

Used the division property of equality.
x   12  3.4641

Solved for x using the square root property and estimated
the solution as a decimal.

Checked the negative solution The positive solution can
also be checked.
1.5 x 2  18
1.5 x 2
1.5
 18
2
x  12
1.5  3.4641  6.1  11.9
2
1.5 12   6.1  11.9
18  6.1  11.9
11.9  11.9
Homework 2.2 Solve and check.
13) a2  36
14) 2x 2  1
17) 3x 2  14  16
18) 4x 2  9  0
Copyright 2014 Scott Storla
15) 9  0.5w 2  9
2.2 Continuing With Operation/Inverse Pairs
16) 1 
k2
 1.4
5
54
2.2.4 Mixed Practice
In this last set of problems I’ll combine the different types of equations. To help classify the
equation pay attention to the last operation, above the basic four, that is being applied to the variable.
Here’s a table showing the names of our equations and their usual number of solutions.
Name
Characteristic
Example
Solutions
Linear
The variable has an exponent of 1.
x 5  2
1
Quadratic
The variable has an exponent of 2.
x2  5  2
2
Square root
The variable is part of the radicand.
x 5  2
1
Logarithmic
The variable is part of the argument of a logarithm.
Exponential
The variable is part of the exponent.
1
ln  x   5  2
1
ex  5  2
1
Before we begin please read this important difference between experts and novices.
Think like an expert
Experts often begin the solution process by classifying the equation,
predicting the number and type of solution(s) and putting together a plan
based on inverses. This, and other techniques, allows experts to
develop a “feel” for their work. Novices on the other hand usually just
start the solution process without any forethought. Before you begin the
solution process classify the equation, predict the number of solutions
you expect and put together a plan for solving the equation.
To help you start processing equations I’d like you to first name the type of equation you have,
decide on the number of solutions you expect and then choose the right process to solve that type of
equation. Build a two column table if you’re uncomfortable with the proper order for the inverse operations.
Homework 2.2 Name the type of equation, discuss the expected number of solutions and put
together a solution process Then solve the equation and check your solution.
19) e x  0.035
23)
ln p
 24  22
2
27) 1  2log t  2
20)
24)
f  1  0.5
1
x  6  1
3
28) 1 et  1  2.359
2
Copyright 2014 Scott Storla
21) 46  2L  24
22) d 2  29  5
25) 4  10x  50  108
26) 7 
29) 2  5 g  1
3
6
2.2 Continuing With Operation/Inverse Pairs
3w 2
2
4
30) 100  1.5y 2  160
55
Homework 2.2
1)
100
2)
1.1275
3)
7.3891
4)
7)
64
8)
16
9)
100
10) 23.04
1
 0.7071
2
15) 0
13) 6
14) 
0.001
16)  12  3.4641
5)
31.6228
11) 4.2436
17) 
6)
0.1353
12) 400
2
 0.8165
3
18) 
3
2
19) It’s an exponential equation so expect one solution. Take the logarithm, base e, of both sides.
3.352
20) It’s a square root equation so expect one solution. Add one to both sides and then square both
sides. 2.25
21) It’s a linear equation so expect one solution. Subtract 24 from both sides and divide by 2. 11
22) It’s a quadratic equation so expect two solutions. Subtract 29, divide by negative 1 and use the
square root property.  24  4.899
23) It’s a logarithmic equation so expect one solution. Add twenty four, multiply by 2 and raise both sides
as log base e. e4  54.5982
24) It’s a linear equation so expect one solution. Subtract six and multiply by 3. 21
25) It’s an exponential equation so expect one solution. Add fifty, divide by negative 4 and take the log of
both sides. log 14.5  1.1614
26) It’s a quadratic equation so expect two solutions. Subtract 2, multiply by four-thirds and use the
20
 2.582 .
3
square root property. 
27) It’s a logarithmic equation so expect one solution. Add two, divide by 2 and then raise to the result
3
as a exponent of 10. 10
2
 31.6228
28) It’s an exponential equation so expect one solution. Subtract one, multiply by two and take the
natural logarithm. 1
29) It’s a square root equation so expect one solution. Add one-sixth, divide by five and square.
1
36
30) It’s a quadratic equation so expect two solutions. Subtract 100, divide by one and a half and use the
square root property.  40  6.3246 .
Copyright 2014 Scott Storla
2.2 Continuing With Operation/Inverse Pairs
56