Solving Log Equations (Part 1)

[Pages:6]16-week Lesson 33 (8-week Lesson 27)

Solving Log Equations (Part 1)

When solving algebraic equations, inverse operations are used to isolate variables.

If 2 + 1 = 3, we would subtract 1 and divide by 2 to get the variable by itself.

3-1 = 2

= 1

If

-2 5

=

8, we would multiply by 5,

then square both sides of the equation,

and finally add 2 to isolate .

= (8 5)2 + 2

= 1602

This idea of using inverses to solve equations continues when solving logarithmic equations; we need to use the inverse of a logarithm in order to solve a logarithmic equation, and that means converting to exponential form.

Example 1: Solve the logarithmic equation log2() = -5 by converting to exponential form. Simplify your answer completely.

To solve a logarithmic equation, convert to exponential form. Remember that a logarithm is simply an exponent, so anything equal to a logarithm is also an exponent. That means that if log2() = -5, then -5 is an exponent.

log2() = -5 converts to = 2-5 because -5 is an exponent

1 = 25

1 = 32

1

16-week Lesson 33 (8-week Lesson 27)

Solving Log Equations (Part 1)

Example 2: Solve each of the following equations by converting to

exponential form, and simplify your answers completely.

a. log8() = 2

b. log() = -2

c. ln() = 0

Once again, to solve each logarithmic equation convert to exponential form. Again, a logarithm is simply an exponent, so anything equal to a logarithm is also an exponent.

log8() = 2 converts to = 82

=

log10() = -2 converts to = 10-2

=

log() = 0 converts to = 0

=

Being able to convert from logarithmic form to exponential form is crucial when solving logarithmic equations. Keep in mind that when converting from one form to the other, THE BASE DOES NOT CHANGE. Base in one form is base in the other form; we simply switch the inputs and outputs because logarithms and exponentials are inverses.

Example 3: Solve each of the following equations by converting to

exponential form, and simplify your answers completely. DO NOT

APPROXIMATE, LEAVE ANSWERS IN EXACT FORM,.

a.

log27()

=

1 3

b. log() = -4

c. ln() = 1

2

16-week Lesson 33 (8-week Lesson 27)

Solving Log Equations (Part 1)

Remember from Lesson 31 when logarithmic functions were introduced that the domain of a logarithmic function such as () = log() is positive numbers only, (0, ). This is because the input is the argument of the logarithm, and since the argument is a power of the positive base ,

it must be positive ONLY.

This is a very important concept that must be kept in mind when solving

logarithmic equations; THE ARGUMENT OF THE LOGARITHM

MUST BE POSITIVE. So when solving logarithmic equations, we must

verify that the argument of the logarithm is positive by taking whatever

solution (or solutions) we get for , and back substituting them into the

original logarithmic equation to verify that they result in a positive

argument. This is similar to what we've done with other equations that

had restricted domains, namely rational equations in Lesson 11

(such

as

1 +1

-

1 -1

=

2)

and

square

root

equations

from

Lesson

15

(such as + 1 = 1 - 2).

Example 4: Solve each of the following equations by converting to

exponential form. Be sure to simplify your answers completely, and be

sure to check your answers. LEAVE ANSWERS IN EXACT FORM,

DO NOT APPROXIMATE.

a.

log25(2

-

)

=

3 2

b.

b. ln(2 + 1) = 0

3

16-week Lesson 33 (8-week Lesson 27)

c.

log (5+1)

2-3

=

2

d.

log(4 - 29) = -1

4 - 29 = -1

4 - 29 = 1

4 - 29 = 1

4 = 1 + 29

= +

= +

e.

log4

( +1 )

3-2

=

-1

2

f.

+1 = 4-12

3-2

+1 3-2

=

1

1

42

+1 = 1

3-2 4

+1 = 1

3-2 2

2( + 1) = 3 - 2

2 + 1 = 3 - 2

=

Solving Log Equations (Part 1)

d.

log27(

-

5)

=

-

2 3

- 5 = 27-23

- 5 =

1

2

273

-

5

=

1 ( 327)2

-

5

=

1 (3)2

-

5

=

1 9

= 1 + 5

9

=

f. ln(2 - 1) = 5

15 - = 8-1

4

15 - = 1

4 8

120 - 22 =

0 = 22 + - 120

0 = (2 - 15)( + 8) 0 = (2 - 15)( + 8)

0 = 2 - 15 ; 0 = + 8

4

16-week Lesson 33 (8-week Lesson 27)

g. ln( + 5) = 1 h. A

log( + 5) = 1

+ 5 = 1

+ 5 =

= -

Solving Log Equations (Part 1)

h. log(2 - 3) = 1

15 - = 8-1

4

15 - = 1

4 8

120 - 22 =

i. ln(4 - 29) = -1

j. H

log(4 - 29) = -1

4 - 29 = -1

4 - 29 = 1

4 - 29 = 1

4 = 1 + 29

= +

=

+

= +

= +

j.

log8

(15

-

)

4

=

-1

15 - = 8-1

4

15 - = 1

4 8

8 (15 - ) = (1) 8

4

8

120 - 22 =

0 = 22 + - 120

0 = 22 - 15 + 16 - 120

0 = (2 - 15)( + 8)

0 = 2 - 15 ; 0 = + 8

= ; = -

5

16-week Lesson 33 (8-week Lesson 27)

Solving Log Equations (Part 1)

Keep in mind that a negative answer or an answer of zero is not necessarily invalid. However an answer that results in a negative argument or an argument of zero is invalid. Regardless of whether your answer is negative, positive, or zero, you must back substitute to see what effect it has on the argument.

Answers to Examples:

1

=

1 32

;

2a.

= 64 ; 2b.

=

1 100

;

2c.

= 1 ;

3a.

= 3 ; 3b.

=

1 10000

; 3c.

= ;

4a.

= -123 ; 4b.

= 0 ; 4c.

=

301 195

;

4d.

=

46 9

;

4e.

= 4 ; 4f.

= 52+1; 4g.

= - 5 ;

4h.

= -2; = 5 ; 4i.

=

1+29 4

;

4j.

=

15 2

,

-8

;

6

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